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\begin{document}
\title{U3\_AOS1\_Topic03-Quadratic functions}
\author[1]{Yanik}%
\affil[1]{ConnectApp}%
\vspace{-1em}
\date{\today}
\begingroup
\let\center\flushleft
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\maketitle
\endgroup
\sloppy
\section*{Question Group information}\label{question-group-information}
\textbf{Name:~} Quadratic functions \textbf{}
\textbf{Summary:}
Quadratics are a group of functions that have a surprising number of
useful characteristics and properties. These can be applied to so many
areas at the VCE level that it's crucial we're familiar with the small
details of these polynomials. First, we'll revise what they look like,
and after we'll explore the different methods of solving them as well as
what their key components are. Our top tip when it comes to quadratics?
Don't forget their details by the end of the course, even though we
cover them early on (this content appears frequently on the exam!)~
\textbf{Videos:}
\textbf{~ ~ Video1 url:}
\textbf{~ ~ Video 1 title:}
\textbf{~ ~ Video 1 thumb url:}
\textbf{Start at:}
\textbf{End at:}
\pagebreak
\textbf{Tutorial number: 1}
~\textbf{Prompt:~}What is quadratic?
\textbf{Title:}~The definition of a quadratic
A quadratic is what we call a \textbf{polynomial.~}A polynomial is an
expression that includes the addition and/or subtraction of variables,
their coefficients and constants; these variables are only allowed to
have a positive whole number powers. In other words, these variables are
only allowed to be in this form:
\par\null\par\null
\begin{center}
$x^n$ \newline
where $n \in N \cup \{0\}$
\end{center}
\par\null
Examples of polynomial includes:
\[x^4+3x^3-x^2+9\]
\[8x^6+7x\]
\[2x+1\]
\[4\]
$4$ is a polynomial because it can be written as $4x^0$
\par\null
On the other hand, these are not polynomials:
\[9x^{-4}+7x^6\]
This expression contains a negative power.
\[-7x^{0.4}+3x^{\frac{2}{3}}\]
This expression contains non-integer powers.
\[\displaystyle \frac{1}{x}\]
This expression can be written as $x^{-1}$, which is in fact a negative power.
Polynomials are usually defined by their highest power. We call the highest power the \textbf{degree}. For example, $x^3 + 2x^2 + 7x - 3$ has a degree of 3. Although it contains x to the power of 0, 1, 2 and 3, the highest power is 3. Hence the degree is 3.
Quadratics are defined as polynomials with a degree of 2. Therefore, examples of quadratics include:
\[x^2+3x+2\]
\[4x^2+14\]
\[2x^2\]
\section*{Quadratics and parabolas --- what's the difference?}
Let's clear the air about quadratics and parabolas. We've heard of the terms quadratics and parabolas by now. What's the difference? They effectively mean the same thing; the word \textbf{quadratic is the name of the function}, and the word \textbf{parabola just describes the general U-shape of a quadratic} when sketched out. We'll use them interchangeably from here on out.
\section*{Domain and range:}
Let's take a look at what the basic quadratic, $y = x^2$, looks like:\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.56\columnwidth]{figures/quadratic-functions-basic-graph/quadratic-functions-basic-graph}
\end{center}
\end{figure}
This is an instance of a positive or U-shaped parabola. If the graph's scales were taller and wider, it would look like the parabola would extend out to the left and right to infinity. We can also say that it would extend upwards to infinity as well.
Now let's take a look at a slightly different looking quadratic, given by the equation $y = -x^2$.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-basic-graph-flipped/quadratic-functions-basic-graph-flipped}
\end{center}
\end{figure}
This example is actually a negative or upside-down parabola. While the overall shape looks similar, the direction of this parabola is the biggest difference. Instead of extending upwards to positive infinity, it in fact reaches down all the way to negative infinity.
These two examples give us a general idea of what the domain and range would be for a quadratic, regardless of its shape. Let's summarise their characteristics here:
\begin{itemize}
\item \textbf{Maximal domain:}
\begin{itemize}
\item[\circ] $x \in R$
\item[\circ] The maximal domain of a quadratic will always be the entire real number set, as we can input any number for $x$ and receive a valid answer, no matter how small or large that number is.
\end{itemize}
\item \textbf{Range:}
\begin{itemize}
\item[\circ] Positive parabola $\Rightarrow y \in [a, \infty)$
\item[\circ] Negative parabola $\Rightarrow y \in (-\infty, a]$
\item[\circ] Here, $a$ represents the $y$ value of the quadratic's turning point.
\item[\circ] Notice how the different directions affect which way the range points.
\end{itemize}
\end{itemize}
\pagebreak
\textbf{Tutorial number: 2}
\textbf{Prompt:~} What can we do with quadratic functions in their
expanded form?
\textbf{Title:~} Quadratic functions in expanded form.
First thing's first, what's the expanded form of a quadratic? Here's what it looks like:
\[y = ax^2 + bx + c\]
In this form, $a$, $b$ and $c$ can take almost every value possible. Specifically, here are their restrictions:
\begin{center}
\begin{align*}
a &\in R \backslash \{0\} \newline
b, c &\in R
\end{align*}
\end{center}
Overall, the form may look somewhat ordinary, yet there is so much we can do with a quadratic function given in this way.
\section*{The $y$-intercept:}
Let's start off with graphing. While this may not be the most friendly form to sketch its graph, we can do something pretty simple to get something quite useful.
Suppose we are asked to sketch the graph of a quadratic, labelling all axial intercepts as well. To do this, we always need to do three things: \newline
\begin{itemize}
\item Find the $x$- and $y$-intercept(s)
\item Find the turning point
\item Sketch the graph itself!
\end{itemize}
The expanded form of a quadratic is particularly useful for finding the $y$-intercept. Let's do this for the following function:
\[f(x) = 3x^2 - 4x + 2\]
Don't forget the process to find the $y$-intercept --- we need to sub in $x = 0$.
\begin{align*}
f(0) &= 3(0)^2 - 4(0) + 2 \newline
f(0) &= 2
\end{align*}
\[\therefore y\text{-intercept at } (0,2)\]
Notice how the $y$-intercept seems to occur at the value of the constant? Maybe that's a coincident? Let's make sure it's not --- we should sub $x = 0$ into the general expanded form from earlier to check this:
\begin{align*}
y &= ax^2 + bx + c \newline
&Sub \; x = 0 \newline
y &= a(0)^2 + b(0) + c \newline
y &= c
\end{align*}
\[\therefore y\text{-intercept at } (0,c)\]
Based on this general example, we can see that the \textbf{$y$-intercept of a quadratic function in the form $y = ax^2 + bx + c$ occurs at $y = c$}. It's a great shortcut that can save time and space by doing less working out.
We can see how this $y$-intercept shortcut looks like on a graph if we sketch out the parabola of $y = f(x)$, where $f(x) = 3x^2 - 4x + 2$ from earlier, on the CAS.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-y-intercept/quadratic-functions-y-intercept}
\end{center}
\end{figure}
\section*{Positive/negative parabolas:}
Another little shortcut relates to how we can tell the general shape of a parabola just by looking at its expanded form. Firstly, there are two types of parabolas that we need to know about:
\begin{enumerate}
\item Positive, U-shaped parabolas.
\item Negative, upside-down parabolas.
\end{enumerate}
Let's take a look at a couple of examples below:\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-parabola-shape/quadratic-functions-parabola-shape}
\end{center}
\end{figure}
The above set of axes shows the graph of two quadratics:
\begin{itemize}
\item Red graph \Rightarrow \; $y = x^2 + 1$
\item Blue graph \Rightarrow \; $y = -x^2 + 1$
\end{itemize}
They're pretty much identical, except for one big thing --- their \textbf{shape/direction}. The blue graph seems to be the upside-down version of the red graph. But how? The answer lies in the form of their equation.
Looking at the two equations, both are in expanded form. It may look slightly different to the original form we hard earlier, $y = ax^2 + bx + c$, but the only thing is that $b = 0$, while $a$ and $c$ take non-zero values.
Let's assign what $a$, $b$ and $c$ are equal to for both graphs, and compare any differences:
\begin{enumerate}
\item Red graph:
\begin{itemize}
\item [\textbullet] $a = 1$
\item [\textbullet] $b = 0$
\item [\textbullet] $c = 1$
\end{itemize}
\item Blue graph:
\begin{itemize}
\item [\textbullet] $a = -1$
\item [\textbullet] $b = 0$
\item [\textbullet] $c = 1$
\end{itemize}
\end{enumerate}
We can see the only difference between the two quadratics is the sign of $a$ --- whether it's positive or negative. \textbf{It's the sign of $a$ that changes the direction of the parabola}, from one that is positive (U-shaped) or negative (upside-down U-shaped). \textit{This works no matter what the values of $b$ and $c$.}
As a summary, here is a break down of how $a$ changes the shape:
\begin{itemize}
\item $a > 0 \Rightarrow$ Positive, U-shaped parabola
\item $a < 0 \Rightarrow$ Negative, upside-down parabola
\end{itemize}
\pagebreak
\par\null\par\null
\textbf{Tutorial number: 3}
\textbf{Prompt:~} What can we do with quadratic functions in their
factorised form? ~
\textbf{Title:~} Quadratic functions in factorised form.~ ~
\par\null\par\null
Let's start off with what the factorised form looks like:
\[y = a(x - m)(x - n)\]
We'll go through each component below, though let's start off with what values $a$, $m$ and $n$ can take:
\begin{center}
\begin{align*}
a &\in R \backslash \{0\} \newline
m, n &\in R
\end{align*}
\end{center}
The factorised form can tell us more information up front than the general form, and is sometimes considered to be a more useful form to have. Let's look into its components.
\section*{Positive/negative parabolas:}
Similar to the general form, the factorised form can also tell us what the overall shape of the parabola will be. This involves looking at the sign of $a$. Let's break this down:
\begin{itemize}
\item $a > 0 \Rightarrow$ Positive, U-shaped parabola
\item $a < 0 \Rightarrow$ Negative, upside
\end{itemize}
Let's take a look at an example to show this difference. Suppose we have the following two functions:
\begin{align*}
f(x) &= (x + 1)(x - 1) \newline
g(x) &= -(x + 1)(x - 1)
\end{align*}
Sketching the graph of $y = f(x)$ and $y = g(x)$ gives us the graphs below.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.63\columnwidth]{figures/quadratic-functions-factorised-parabola-shape/quadratic-functions-factorised-parabola-shape}
\end{center}
\end{figure}
From here, we can see that the negative sign in front of the factors flips the graph from $y = f(x)$ to $y = g(x)$. This is a visualisation of the two cases for $a$, where $a > 0$ and $a < 0$.
One thing to notice is that by adding a negative in front of $f(x)$ to transform it into $g(x)$ we have reflected the graph in the $x$-axis, not on its turning point. The reason for this is covered specifically in the transformations topic.
\section*{$x$-intercepts:}
The most helpful piece of information from this form is the location of all the $x$-intercepts. This is especially useful when sketching the graph without having to do any working out.
Put simply, the $x$-intercepts are located at:
\begin{center}
$(m, 0) \; \text{and} \; (n, 0)$
\end{center}
Why is this? Let's take a look at how we usually solve for $x$-intercepts. By letting $y = 0$, we can find the $x$-intercepts for the overall factorised form.
\begin{align*}
y &= a(x - m)(x - n) \newline
&Sub \; y = 0 \newline
0 &= a(x - m)(x - n) \newline
0 &= (x - m)(x - n)
\end{align*}
It should be noted that the $a$ term can be taken to the other side by dividing both sides by $a$. This means that we don't have to deal with an extra term when solving.
Using the null factor law, we can break this up into two solutions:
\begin{align*}
x - m &= 0 \; \text{or} \; x - n = 0 \newline
\Rightarrow \; x &= m \; \text{or} \; n \newline
\newline
\therefore \; x&\text{--intercepts at} \; (m, 0) \; \text{and} \; (n, 0)
\end{align*}
What does this look like on a graph? Let's take the following quadratic function as an example:
\[y = (x + 1)(x - 2)\]
Notice how one factor has a plus sign, while the other has a minus sign. Looking back at our factorised form, it says that both factors have a minus sign. Does this pose a problem to us if we were to find the $x$-intercepts? Not at all! We just need to know how to handle the two different signs.
First of all, let's identify what $m$ and $n$ are in our example:
\begin{itemize}
\item $m = -1$
\item $n = 2$
\end{itemize}
The question is how did we get $m = -1$ when it's clearly $(x + 1)$ in the equation? Let's take it step by step. Say that we let $m = 1$ and $n = 2$, and then sub these into our factorised form. Here's what we would get:
\begin{align*}
y &= (x - m)(x - n) \newline
&Sub \; m = 1 \; and \; n = 2 \newline
\Rightarrow y &= (x - 1)(x - 2)
\end{align*}
We end up having $(x - 1)$ instead of $(x + 1)$. The concept we've come across is the opposite sign rule.
The \textbf{opposite sign rule} states that to get the $x$-intercept from the factorised form, we need to swap the sign in front of the number. This also works the other way, meaning to get the factor from the $x$-intercept, we only have to swap the sign in front of the number.
Using this, let's try substituting in our original two values, $m = -1$ and $n = 2$:
\begin{align*}
y &= (x - m)(x - n) \newline
&Sub \; m = -1 \; and \; n = 2 \newline
y &= (x - (-1))(x - 2) \newline
y &= (x + 1)(x - 2)
\end{align*}
And it works! The opposite sign rule also comes in handy when dealing with translations under the transformation topic --- this will be covered in a later topic.
Using this, we can sketch the graph based on the two intercepts and the fact that it's a positive quadratic.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.56\columnwidth]{figures/quadratic-functions-factorised-x-intercepts/quadratic-functions-factorised-x-intercepts}
\end{center}
\end{figure}
\subsection*{Special cases:}
There are some instances when particular values of $m$ and $n$ produce interesting results. Let's list them out here before we go into detail.
\begin{enumerate}
\item $m = n$
\item $m = -n$
\end{enumerate}
\newline
\textbf{1. $m = n$}
What does it mean when $m = n$? Let's take a look by substituting it into our factorised form.
\begin{align*}
y &= a(x - m)(x - n) \newline
&Sub \; m = n \newline
y &= a(x - m)(x - m) \newline
\Rightarrow y &= a(x - m)^2
\end{align*}
What does this new equation mean? First of all, this is a quadratic factor, meaning it is a set of squared brackets. It still tells us about the $x$-intercepts since it's still in factorised form. However, if we try to find the $x$-intercepts by hand we'll notice that we only get one solution, which implies that there is only one $x$-intercept.
\begin{align*}
y &= a(x - m)^2 \newline
&Sub \; y = 0 \newline
0 &= a(x - m)^2 \newline
\Rightarrow x - m &= 0 \newline
\Rightarrow x &= m
\end{align*}
This working out tells us that there is only one $x$-intercept at $(m, 0)$. Let's suppose that $m = 2$ such that our function is $y = (x - 2)^2$ --- here's what it would look like.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-factorised-one-x-intercept/quadratic-functions-factorised-one-x-intercept}
\end{center}
\end{figure}
\textbf{2. $m = -n$}
While it looks somewhat similar to the previous case, $m = -n$ produces a different special case. Let's plug this into our factorised form and see what we get.
\begin{align*}
y &= a(x - m)(x - n) \newline
y &= a(x - m)(x + (-n)) \newline
&Sub \; m = -n \newline
\Rightarrow y &= a(x - m)(x + m) \newline
\end{align*}
While it may not look so special right now, we can notice a new characteristics if we go to find its $x$-intercepts.
\begin{align*}
y &= a(x - m)(x + m) \newline
&Sub \; y = 0 \newline
0 &= a(x - m)(x + m) \newline
\Rightarrow x - m &= 0 \; or \; x + m = 0 \newline
x &= -m \; or \; m \newline
\Rightarrow x &= \pm m
\end{align*}
We ended up with the positive and negative solution of the same number! The significance of this result will make more sense when we sketch out a graph. Let's consider the quadratic $y = (x + 2)(x - 2)$.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.56\columnwidth]{figures/quadratic-functions-factorised-symmetrical/quadratic-functions-factorised-symmetrical}
\end{center}
\end{figure}
The biggest thing to notice is how both $x$-intercepts are symmetrical about the $y$-axis. This is true with any value of $m$, as long as the quadratic is in the form $y = a(x + m)(x - m)$.
\pagebreak
\par\null
\textbf{Tutorial number: 4}
\textbf{Prompt:~} What can we do with quadratic functions in their
turning point form? ~
\textbf{Title:~} Quadratic functions in turning point form.~ ~
\par\null
Turning point form can sometimes be all you need to successfully sketch the \textbf{shape} of the parabola given. Its usefulness comes from a certain piece of information that no other form can provide. We'll see what this is soon!
Before then, let's begin with what the turning point form looks like:
\[y = a(x - h)^2 + k\]
In particular, here are the values that each constant can take:
\begin{center}
\begin{align*}
a &\in R \backslash \{0\} \newline
h, k &\in R
\end{align*}
\end{center}
Let's get straight into the usefulness of this form.
\section*{Positive/negative parabolas:}
As we have seen with the previous two forms, turning point form can also tell us what the general \textbf{shape and direction} of the parabola is. This is dependent on $a$, and the following tells us what the two different scenarios are:
\begin{itemize}
\item $a > 0 \Rightarrow$ Positive, U-shaped parabola
\item $a < 0 \Rightarrow$ Negative, upside-down parabola
\end{itemize}
We'll explore how this changes the shape in the examples found in the next section.
\section*{The turning point:}
As the name suggests, the most important piece of information this form provides is the location of the turning point. How uncanny!
If we have a quadratic in turning point form, we can obtain the coordinates of the turning point (TP) as follows:
\begin{center}
$y = a(x - h)^2 + k$
\end{center}
\begin{center}
$\Rightarrow TP \; \text{at} \; (h, k)$
\end{center}
As we similarly covered in the factorised form, the opposite sign rule also applies here with anything next to the $x$. The following example illustrates this.
Consider the following quadratics:
\begin{align*}
y &= (x - 3)^2 - 1 \qquad (1) \newline
y &= (x + 3)^2 - 1 \qquad (2)
\end{align*}
Equation $(1)$ has a turning point at $(3, -1)$, whereas equation $(2)$ has a turning point at $(-3, -1)$. This means that we must swap the sign for the number next to $x$ when finding the $x$ value of the turning point. In other words, we need to apply the opposite axis rule here as well. As for the $y$ value of the turning point, this is just the constant at the end of the equation --- the sign remains the same.
\subsection*{Changes in $h$ and $k$:}
Let's investigate how different values of $h$ affect a quadratic graph.
Let's take a look at the difference between equations $(1)$ and $(2)$ on a graph. We've also sketched the graph of $y = x^2$ to compare the changes from the basic parabola.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-tp-opposite-sign/quadratic-functions-tp-opposite-sign}
\end{center}
\end{figure}
Let's categorise the differences:
\begin{itemize}
\item Effect of $h$:
\begin{itemize}
\item[\circ] We can see that the sign of what's inside the bracket changes whether the turning point is to the left or the right of the $y$-axis.
\item[\circ] In other words: if there is a minus sign on the inside, the turning point is on the right; whereas if there is a plus sign on the inside, the turning point is on the left.
\end{itemize}
\item Effect of $k$:
\begin{itemize}
\item[\circ] Both graphs have the $y$ value of their turning points located at $-1$; this corresponds to the $-1$ constant that's found at the end of both equations.
\item[\circ] If there was a $+1$ at the end of the both equations, the graphs' turning points would be located above the axis, with a $y$ value of $1$.
\end{itemize}
\item Relative to $y = x^2$:
\begin{itemize}
\item[\circ] The basic graph of $y = x^2$ can be expressed in turning point form with $a = 1$, $h = 0$ and $k = 0$.
\item[\circ] We can see that the graphs move away from the basic graph as $h$ and $k$ change.
\end{itemize}
\end{itemize}
While this form is handy when interpreting characteristics of quadratics, it is not limited to only this. This form can also be used in transformations, which is covered in a later topic, where we'll specifically explain what effect each of $a$, $h$ and $k$ have on the graph.
\pagebreak
\par\null
\textbf{Tutorial number: 5}
\textbf{Prompt:~} What can we do with quadratic functions in any of
their forms?
\textbf{Title:~}~Summary of quadratic functions in all their forms.
We've been through three different forms in which we can be given quadratic functions. It's definitely a lot to take in, with each one having many characteristics, so let's summarise them all into a neat package we can revisit.
\section*{Summary of quadratic forms:}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-forms-summary/quadratic-functions-forms-summary}
\end{center}
\end{figure}
The above table describes all the crucial characteristics that each form tells us. Each form has its own unique information, though all of them also tell us about the shape and direction of the quadratic.
\subsection*{Effects of $a$:}
One thing that has only been lightly touched on so far is the effect of $a$ in each of the graphs. We have already discussed the effect whether $a$ is positive or negative, and how this flips the shape of the graph, though we have not talked about \textbf{how the magnitude or size of $a$ affects the graph}. We'll discuss this here, though the general effect of $a$ will also be thoroughly discussed in the transformations topic.
First of all, there is some notation used in the table that has yet to be covered. The two vertical bars surrounding the $a$ are referred to as the ``\textbf{magnitude}''. When written like as $|a|$, the expression is read as ``the magnitude of $a$''. Here's all we need to know about the magnitude function in Methods:
\begin{itemize}
\item If the number inside the magnitude function is positive, then the magnitude is the positive number.
\item If the number inside the magnitude function is negative, then the magnitude is the number without the minus sign.
\end{itemize}
Put simply, \textbf{the magnitude of any number is just the positive number}, even if it's already positive. Here's a few examples:
\begin{center}
\begin{align*}
|1| &= 1 \newline
|-1| &= 1 \newline
|3-1| &= |2| = 2 \newline
|1-3| &= |-2| = 2
\end{align*}
\end{center}
Let's start talking about how the magnitude of $a$ affects the shape of a parabola.
\begin{itemize}
\item $|a| > 1$:
\begin{itemize}
\item[\circ] For example: $2$, $-4$, $15$, or $-100$.
\item[\circ] If the size of $a$ is more than $1$, then the graph is being \textbf{stretched} vertically.
\item[\circ] Imagine grabbing the parabola from the top and the bottom, and stretching it vertically away from the $x$-axis.
\item[\circ] In the end, we're left with a \textbf{narrower or skinnier} parabola.
\end{itemize}
\item $|a| < 1$:
\begin{itemize}
\item[\circ] For example: $\displaystyle \frac{1}{2}$, $\displaystyle -\frac{3}{5}$, $0.45$, or $-0.24$
\item[\circ] If the size of $a$ is less than $1$, then this means the graph is being \textbf{squashed} vertically.
\item[\circ] Imagine grabbing the parabola from the top and the bottom, then squashing it vertically towards the $x$-axis.
\item[\circ] In the end, we're left with a \textbf{wider or fatter} parabola.
\end{itemize}
\end{itemize}
Some examples below explain how changes in $a$ affect the graph.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-dilations/quadratic-functions-dilations}
\end{center}
\end{figure}
\subsection*{Getting between different forms:}
While it's great to know what each form tells us, we should also know how to convert between two different forms so that we can extract all the information we can. The following diagram broadly exlpains the methods to convert between the three forms.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratic-functions-forms-conversion/quadratic-functions-forms-conversion}
\end{center}
\end{figure}
\begin{enumerate}
\item From \textbf{expanded form} to \textbf{factorised form}:
\begin{itemize}
\item[\textbullet] Any methods of factorisation can be used here; the method covered in algebraic techniques that is the most convenient should be used.
\end{itemize}
\item From \textbf{expanded form} to \textbf{turning point form}:
\begin{itemize}
\item[\textbullet] The completing the square technique needs to be specifically used here. This was covered in the algebraic techniques topic.
\end{itemize}
\item From \textbf{factorised form} to \textbf{expanded form}:
\begin{itemize}
\item[\textbullet] We only need to expand the two linear factors out to reach the expanded form.
\item[\textbullet] Don't forget to account for the $a$ that may be at the front of the expression.
\end{itemize}
\item From \textbf{turning point form} to \textbf{factorised form}:
\begin{itemize}
\item[\textbullet] We need to specifically use the difference of two squares technique here. This was covered in the algebraic techniques topic.
\item[\textbullet] This will only work if either $a$ or $k$ is negative; they cannot be both positive or negative at the same time since they will no longer be a ``difference'' of two squares.
\end{itemize}
\item From \textbf{turning point form} to \textbf{expanded form}:
\begin{itemize}
\item[\textbullet] We simply need to expand the quadratic factor according to the expansion rules in the algebraic techniques topic.
\item[\textbullet] Don't forget to multiply by the $a$ at the front, and add the $k$ at the very end.
\end{itemize}
\end{enumerate}
\pagebreak
\textbf{Tutorial number: 6}
\textbf{Prompt:~} What other information can we get from the quadratic
formula?~ ~~
\textbf{Title:~} Components of the quadratic formula.~ ~
\par\null
The quadratic formula is given by:
\[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
This formula is roughly made up by two parts:
\[-\frac{b}{2a}\]
\[b^2-4ac\]
Both of them have very useful meaning and applications. Let's have a look at them separately.
\subsection*{The turning point formula}
The first part of the quadratic formula actually gives us the $x$ value of the turning point of any quadratic graphs.
\[x=-\frac{b}{2a}\]
Hence, to find where the turning point is located when given an equation in general form, this is one of the fastest way to obtain that.
How do we get the $y$ value of the turning point?
There are two ways to do it:
\begin{itemize}
\item using another formula $\displaystyle y=c-\frac{b^2}{4a}$
\item substituting the $x$ value of turning point into the equation to get the $y$ value
\end{itemize}
Some people prefer the second way which is simply substituting the $x$ value into the equation as it is the most intuitive and there will be one less formula to memorise. Again, it is completely up to you.
For example, where is the turning point of a parabola $y=x^2+4x-32$?
In this equation, we know that $a=1$, $b=4$ and $c=-32$. Putting these numbers into the turning point formula:
\begin{align*}
\newline
\displaystyle x&= -\frac{b}{2a}
\newline
\displaystyle x&= -\frac{4}{2(1)}
\newline
\displaystyle x&= -2
\end{align*}
The turning point occurs at $x=-2$.
Now, let's find the $y$ value of the turning point. We can use both methods. Let's try using the formula.
\begin{align*}
\newline
\displaystyle y&=c-\frac{b^2}{4a}
\newline
\displaystyle y&=-32-\frac{4^2}{4(1)}
\newline
\displaystyle x&= -32-4
\newline
\displaystyle x&= -36
\end{align*}
By substituting $x=-2$ into $y=x^2+4x-32$.
\begin{align*}
\newline
\displaystyle y&=(-2)^2+4(-2)-32
\newline
\displaystyle y&=4-8-32
\newline
\displaystyle x&= -36
\end{align*}
So the turning point is at $(-2,-36)$.
\subsection*{The discriminant}
The discriminant is given by:
\[\Delta=b^2-4ac\]
where $a$, $b$ and $c$ are from $0=ax^2+bx+c$.
But what is the discriminant for?
The discriminant tells us how many solutions there are in a quadratic equation. How? Well, take a look at where the $b^2-4ac$ is positioned in the quadratic formula. It's under the surd (the square root)! Something you should remember is that you can't find the square root of a negative number, so if this expression is less than zero, there are no real solutions. On the other hand, if the $b^2-4ac$ is a positive number, we can find a positive and negative square root, and this results in two solutions for the $x$-intercept. But what if the discriminant is zero? Well, the square root of zero is just zero! So, in the quadratic formula you'd be adding and subtracting zero -- in other words, you'd just have one solution instead of two. In summary:
\begin{itemize}
\item When $\Delta<0$, there are no real solutions
\item When $\Delta=0$, there is one real solution
\item When $\Delta>0$, there are two real solutions
\end{itemize}
For example, we know that $x^2 + 5x+6=0$ has two solutions, which are $x=-2$ and $x=-3$. Let's look at the discriminant for this equation. Since $a=1$, $b=5$ and $c=6$,
\begin{align*}
\newline
\Delta&=b^2-4ac
\newline
\Delta&=5^2-4(1)(6)
\newline
\Delta&=25-24
\newline
\Delta&=1
\end{align*}
Since $\Delta>0$, it is indicating this equation has two real solutions.
Let's look at another example, $x^2-14+49=0$.
\begin{align*}
\newline
\Delta&=b^2-4ac
\newline
\Delta&=14^2-4(1)(49)
\newline
\Delta&=196-196
\newline
\Delta&=0
\end{align*}
Since $\Delta=0$, it is indicating this equation only has one real solutions. Let's try to solve this and see if this is true.
\begin{align*}
\newline
x^2-14+49=0
\newline
(x-7)^2=0
\newline
(x-7)=0
\newline
x=7
\end{align*}
By solving the equation, we can see that there is only one solution and it is $x=7$.
How about $y=x^2-3x+7$?
\begin{align*}
\newline
\Delta&=b^2-4ac
\newline
\Delta&=(-3)^2-4(1)(7)
\newline
\Delta&=9-28
\newline
\Delta&=-19
\end{align*}
From the discriminant, we can see that there are no real solutions.
This is true, because if we cannot factorise this, even using completing the square. Feel free to give it a try!
~
\pagebreak
\section*{Question 1:~}\label{question-1}
\textbf{Title:~}Identify which of the following is a quadratic
expression.
\textbf{Tags:} polynomial, quadratic
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
\textbf{Prompt:}
\par\null
The expression that represents a quadratic polynomial is:
\par\null
A:~~
$x^2 + 3x +4x^5$
~ ~ Explanation:
Not quite! Although all the powers are positive, and even a power of 2
exists, the fact that there's also a power of 5 means that the degree is
now 5 rather than 2. Hence, this polynomial is in fact a quintic and not
a quadratic.
\par\null
B:
$\displaystyle x^2 + \frac{3}{x^2} +9$
~ ~ Explanation:
\par\null
This isn't quite right. Think back to the definition of a quadratic ---
it must be a polynomial of degree 2. While there is a power of 2 in the
expression, representing a quadratic term, the expression itself isn't
actually a polynomial to begin with.
$\displaystyle \frac{3}{x^2}$ can be written as $3x^{-2}$. Since there's a negative power here, this expression is not a polynomial, and therefore it can't be a quadratic.
\par\null
C:~
$-6x^2+8x-10$
~ ~ Explanation:
That's it! A quadratic is a polynomial with degree of 2.~ This means the
highest power of the polynomial must be 2.~
\par\null
This expression is definitely a polynomial since all its powers are non-negative whole numbers (don't forget that $10$ is the same as $10x^0$). The highest power in the expression is 2, which then makes it a quadratic.
\par\null
D:
$5x+1$
~ ~ Explanation:
{That's not quite right, though you're halfway there! For an expression
to be a quadratic, it must be a polynomial with a degree of 2.}
\par\null
{Although all the powers are non-negative whole numbers, which makes it
a polynomial, the highest power is only 1. As such this expression can't
be called a quadratic, but rather a linear polynomial.}
\section*{Question 1 Hint Menu}\label{question-1-hint-menu}
Copy/paste the following tutorials:
\begin{itemize}
\tightlist
\item
\emph{~ ~ ~ ~ The definition of a quadratic}
\end{itemize}
\textbf{Question analysis}
There are two components that make up a quadratic: it must be a
polynomial; and it must have a degree of 2. Understanding this
definition and how to look for its characteristics should help you to
answer this question.
~
\pagebreak
\section*{Question 2:~}\label{question-2}
\textbf{Title:~}Identify an equation that has no solutions.
\textbf{Tags:~}quadratic, discriminant
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
\textbf{Prompt:}
\par\null
The quadratic equation that has no solution for $x$ is:
\par\null
A:~ (correct)
$-6x^2 +7x-3=0$
~ ~ Explanation:
Yes, that's right! If we're asked to see if a quadratic equation has any solutions we should go straight to using the discriminant. This is arguably the quickest method to working it out.
\begin{align*}
\newline
\Delta&=b^2-4ac
\newline
\Delta&=7^2-4(-6)(-3)
\newline
\Delta&=49-72
\newline
\Delta&=-23
\end{align*}
Since $\Delta<0$, there are no solutions.
\par\null
B:
$x^2 +8x+7=0$
~ ~ Explanation:
This isn't the right answer, unfortunately. Let's go back to basics. If a question is asking us about how many solutions a quadratic equation has, there's a quick method that allows us to do figure this out, and that's the discriminant. Don't forget that the value of the discriminant can tell us how many solutions the equation will have.
\begin{align*}
\Delta &> 0 \Rightarrow \; \text{Two solutions} \newline
\Delta &= 0 \Rightarrow \; \text{One solution} \newline
\Delta &< 0 \Rightarrow \; \text{No solutions}
\end{align*}
Let's use the discriminant to work out if this equation has no solutions.
\begin{align*}
\newline
\Delta&=b^2-4ac
\newline
\Delta&=8^2-4(1)(7)
\newline
\Delta&=64-28
\newline
\Delta&=36
\end{align*}
Since $\Delta>0$, this equation has two solutions.
\par\null
C:
$x^2-2=0$
~ ~ Explanation:
This isn't quite right. Let's go back to basics. If a question is asking us about how many solutions a quadratic equation has, there's a quick method that allows us to do figure this out, and that's the discriminant. Don't forget that the value of the discriminant can tell us how many solutions the equation will have.
\begin{align*}
\Delta &> 0 \Rightarrow \; \text{Two solutions} \newline
\Delta &= 0 \Rightarrow \; \text{One solution} \newline
\Delta &< 0 \Rightarrow \; \text{No solutions}
\end{align*}
Let's use the discriminant to work out if this equation has no solutions.
\begin{align*}
\newline
\Delta&=b^2-4ac
\newline
\Delta&=0^2-4(1)(-2)
\newline
\Delta&=0+8
\newline
\Delta&=8
\end{align*}
Since $\Delta>0$, this equation has two solutions.
\par\null
D:
$(x-3)^2=0$
~ ~ Explanation:
This answer isn't quite correct. Let's go back to basics. If a question is asking us about how many solutions a quadratic equation has, there's a quick method that allows us to do figure this out, and that's the discriminant. Don't forget that the value of the discriminant can tell us how many solutions the equation will have.
\begin{align*}
\Delta &> 0 \Rightarrow \; \text{Two solutions} \newline
\Delta &= 0 \Rightarrow \; \text{One solution} \newline
\Delta &< 0 \Rightarrow \; \text{No solutions}
\end{align*}
To use the discriminant for this equation, though, we need to expand the quadratic factor out to obtain the expanded form of $y = ax^2 + bx + c$. This is the form that the discriminant's constants come from.
\[(x-3)^2=x^2-6x+9\]
Now that we have this, let's use the discriminant to work out if this equation has no solutions.
\begin{align*}
\newline
\Delta&=b^2-4ac
\newline
\Delta&=(-6)^2-4(1)(9)
\newline
\Delta&=36-36
\newline
\Delta&=0
\end{align*}
Since $\Delta=0$, this equation actually has only one solution.
\par\null
\section*{Question 2 Hint Menu}\label{question-2-hint-menu}
Copy/paste the following tutorials:
\begin{itemize}
\tightlist
\item
\emph{Components of the quadratic formula}
\end{itemize}
\textbf{Question analysis}
This question is asking us to find which one of the equations has no
solutions. Typically, we could try to solve the equations to see if they
have any solutions. However, it doesn't make sense to solve an equation
if we expect zero solutions. Think back to thetools we have available to
work out the~\textbf{number~}of solutions that a quadratic equation will
have, rather than solving the equation itself.
\par\null\par\null\par\null
~
\pagebreak
\section*{Question 3:~}\label{question-3}
\textbf{Title:~}Calculate the coordinates of the turning point for a
quadratic equation.
\textbf{Tags:} quadratic, turning point
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
\textbf{Prompt:}
\par\null
The coordinates of the turning point for the quadratic $y = 2x^2+5x-1$ are:
\par\null
A:
\par\null
$\displaystyle (-\frac{25}{4}, -\frac{29}{4})$
~ ~ Explanation:
This isn't quite right. If you tried to convert the quadratic into turning point form, that's great thinking! It gives us the turning point straight away which is really handy. However, you may have made some arithmetic errors in your working out, possibly when completing the square.
Let's have a look at how to start completing the square for this quadratic. Don't forget that we always have to factorise out any coefficients on $x^2$ before starting (this is probably what you forgot!).
\begin{align*}
y &= 2x^2 + 5x - 1 \newline
y &= 2(x^2 + \frac{5}{2}x - \frac{1}{2}) \newline
\textbf{CHSS} \Rightarrow \quad y &= 2[(x^2 + \frac{5}{2}x + \frac{25}{16}) - \frac{25}{16} - \frac{1}{2}]
\end{align*}
From here on out, we have to identify the perfect square so that we can finish completing the square. Give it another crack now that we've started out on the right foot!
\par\null
B:
\par\null
$\displaystyle (-\frac{5}{4}, -\frac{33}{16})$
~ ~ Explanation:
That's not quite right, though you're very close! If you tried to convert the quadratic into turning point form, thats great thinking! It gives us the turning point straight away which is really handy. However, you may have made some calculation errors in your working out, possibly when finishing completing the square.
Let's have a look at how we would complete the square on this quadratic. Don't forget that we always have to factorise out any coefficients on $x^2$ before starting.
\begin{align*}
y &= 2x^2 + 5x - 1 \newline
y &= 2(x^2 + \frac{5}{2}x - \frac{1}{2}) \newline
\textbf{CHSS} \Rightarrow \quad y &= 2[(x^2 + \frac{5}{2}x + \frac{25}{16}) - \frac{25}{16} - \frac{1}{2}] \newline
y &= 2[(x + \frac{5}{4})^2 - \frac{25}{16} - \frac{8}{16}] \newline
y &= 2[(x + \frac{5}{4})^2 - \frac{33}{16}]
\end{align*}
We can't stop here, though! Think back to the general turning point form --- it doesn't include any extra brackets around the quadratic term. Hence, we can't use the last line of our working out to read the turning point. While the $x$ value is correct, the $y$ value still needs some work.
To get our equation to the turning point form, we just need to expand the $2$ back into the rest of the expression. Once we've done that, we should get the correct answer!
\par\null
C: (correct)
\par\null
$\displaystyle (-\frac{5}{4}, -\frac{33}{8})$
~ ~ Explanation:
That's correct! We're asked to find the coordinates of the turning point for a quadratic that's in expanded form. Our goal is to find the most efficient way to get the turning point from this form. The best tool that we have is the formula for the $x$ value of the turning point:
\begin{align*}
x &= -\frac{b}{2a} \newline
x &= -\frac{5}{2(2)} \newline
x &= -\frac{5}{4}
\end{align*}
It should be noted that we can't stop here --- the question has specifically asked us for a coordinate, meaning we also need the $y$ value. While this point doesn't have much significance here, \textbf{in short answer question students often get caught out leaving their answer as an $x$ value}, rather than an $x$ and $y$ coordinate pair.
On that note, let's find that $y$ value. While we did cover a formula to calculate the $y$ value of the turning point, we will simply substitute the $x$ value back into the equation. Why do this? It emphasises the fact that, by subbing in $x$ value into the original equation, we obtain its respective $y$ value.
\begin{align*}
&Sub \; x = -\frac{5}{4} \; into \; y = 2x^2 + 5x - 1 \newline
y &= 2(-\frac{5}{4})^2 + 5(-\frac{5}{4}) - 1 \newline
y &= 2(\frac{25}{16}) - \frac{25}{4} - 1 \newline
y &= \frac{25}{8} - \frac{50}{8} - \frac{8}{8} \newline
\Rightarrow y &= -\frac{33}{8}
\end{align*}
Now that we have the $x$ and $y$ value, we can express them as a coordinate:
\[(-\frac{5}{4}, -\frac{33}{8})\]
\par\null
D:
\par\null
$\displaystyle (-\frac{5}{4}, -\frac{26}{8})$
\par\null
~ ~ Explanation:
That's so close but not quite there! It's just the $y$ value that's slightly off. This could be due to a calculation error in your working out. You started out right, since the $x$ value you got from using the turning point formula is correct:
\begin{align*}
x &= -\frac{b}{2a} \newline
x &= -\frac{5}{2(2)} \newline
x &= -\frac{5}{4}
\end{align*}
It's during the next step - subbing the $x$ value into the equation to get the $y$ value - that you probably made an error. The following working out shows how we may have reached our incorrect answer.
\begin{align*}
&Sub \; x = -\frac{5}{4} \; into \; y = 2x^2 + 5x - 1 \newline
y &= 2(-\frac{5}{4})^2 + 5(-\frac{5}{4}) - 1 \newline
y &= 2(\frac{25}{16}) - \frac{25}{4} - 1 \newline
y &= \frac{25}{8} - \frac{50}{8} - 1 \newline
y &= -\frac{25}{8} - 1 \newline
\Rightarrow y &\neq -\frac{26}{8}
\end{align*}
The last two steps in our $y$ value working out is where the error may have come from. Scan back through those steps and see what we can do to fix it up. Don't forget that when subtracting fractions, both terms must be expressed as a fraction with a common denominator before we can subtract the numerators.
\par\null\par\null\par\null
\section*{Question 3 Hint Menu}\label{question-3-hint-menu}
Copy/paste the following tutorials:
\begin{itemize}
\tightlist
\item
\emph{Components of the quadratic formula}
\item
\emph{Quadratic functions in turning point form}
\end{itemize}
\textbf{Question analysis}
\par\null
This question is asking us to find the coordinates of a quadratic's turning point. To do this we need both the $x$ and $y$ value to write the coordinate. Usually with coordinates, we try to find the $x$ value first. The only issue is what methods and tools do we have to find the $x$ value of the turning point? Think back to the various forms quadratics can be expressed in (expanded, factorised, turning point form)... which one is most useful? Alternatively, what formulas have we learnt that enable us to calculate the $x$ value of the turning point?
Once we have the $x$ value, how do we find the $y$ value? All we have to do is substitute the $x$ value straight back into our original function's equation.
\pagebreak
\section*{Question 4:~}\label{question-4}
\textbf{Title:~}Identify a graph that represents a quadratic function.
\textbf{Tags:} quadratic, factorised form, factorised, parabola, graph
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
\textbf{Prompt:}
\par\null
Consider the quadratic function $y = -(x-3)(1-x)$.
The graph that best represents this function is:
\par\null
A:
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.42\columnwidth]{figures/quadratics-mcq04-1/quadratics-mcq04-1}
\end{center}
\end{figure}
~ ~ Explanation:
This graph is really close, though it's not quite right! You've got the correct $x$-intercepts, but the issue is in the general direction of the parabola. Let's take a closer look at the equation...
The quadratic seems to be in factorised form, though on further inspection the $(1 - x)$ factor looks slightly out of place. To properly read it in factorised form, we need to change $-x$ into $x$. How would we do that?
\begin{align*}
y &= -(x - 3)(1 - x) \newline
\Rightarrow y &= -(x - 3)[-(x - 1)]
\end{align*}
Looking at this, we only have to factorise a $-1$ out of that factor to transform $-x$ to $x$. By simplifying, we'll get the correct answer.
Now that we have two factors next to each other, we can read what the $x$-intercepts are. Using our opposite sign rule, we can see that the factor $(x - 3)$ corresponds to an $x$-intercept at $(3, 0)$, and the factor $(x - 1)$ corresponds to $(1, 0)$; this matches what we can see on the graph.
To find the general direction of the graph, whether it's positive or negative, we need to check the value of $a$. Since there are two minus signs in front of the two factors, we can't tell this value straightaway --- we need to do a bit of manipulation. Because they are multiplied together, they actually cancel out to become a positive sign. Thus, the entire quadratic has a positive $a$ value. Thinking back to the characteristics of the factorised form, what does a positive value of $a$ tell us about its direction?
\par\null
B:
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.42\columnwidth]{figures/quadratics-mcq04-2/quadratics-mcq04-2}
\end{center}
\end{figure}
~ ~ Explanation:
That's not quite right! The general direction (negative and upside-down) of this graph doesn't actually match what the direction of this quadratic should be. You might have gotten confused with the many minus signs in the equation. The $x$-intercepts aren't quite right either. Let's take a closer look at the equation...
The quadratic seems to be in factorised form, though on further inspection the $(1 - x)$ factor looks slightly out of place. To properly read it in factorised form, we need to change $-x$ into $x$. How would we do that?
\begin{align*}
y &= -(x - 3)(1 - x) \newline
\Rightarrow y &= -(x - 3)[-(x - 1)]
\end{align*}
Looking at this, we only have to factorise a $-1$ out of that factor to transform $-x$ to $x$. By simplifying, we'll get the correct answer.
Since there are two minus signs multiplied together, they actually cancel out to become a positive sign. Thus, the entire quadratic has a positive $a$ value. Thinking back to the characteristics of the factorised form, what does a positive value of $a$ tell us about its direction?
Now let's address the $x$ intercepts. Looking at the equation, we can see that one of the factors is $(x - 3)$. Don't forget that the opposite sign rule applies here: we need to switch the sign in front of the number inside the factor to get the $x$-intercept. This means that one of the intercepts should be at $(3, 0)$, whereas this graph shows an intercept at $(-3, 0)$. Try to simplify the rest of the equation and apply this process to the second factor!
\par\null
C: (correct)
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.42\columnwidth]{figures/quadratics-mcq04-3-correct/quadratics-mcq04-3-correct}
\end{center}
\end{figure}
~ ~ Explanation:
That's correct --- great work!
The easiest way to get the graph from the function is to look at the equation's form and its characteristics. The quadratic seems to be in factorised form, though on further inspection the $(1 - x)$ factor looks slightly out of place. To properly read it in factorised form, we need to change $-x$ into $x$. How would we do that?
\begin{align*}
y &= -(x - 3)(1 - x) \newline
y &= -(x - 3)[-(x - 1)] \newline
y &= (x - 3)(x - 1) \newline
\Rightarrow y &= (x - 1)(x - 3)
\end{align*}
Looking at this, we only have to factorise a $-1$ out of that factor to transform $-x$ to $x$. Now that it's properly in factorised form, we can see the following characteristics:
\begin{itemize}
\item $a > 0 \Rightarrow$ Positive, U-shaped parabola.
\item $(x - 1) \Rightarrow x$-intercept at $(1, 0)$
\item $(x - 3) \Rightarrow x$-intercept at $(3, 0)$
\end{itemize}
From all this, the graph in this option is the closest one matching this description.
\par\null
D:
\par\null\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.42\columnwidth]{figures/quadratics-mcq04-4/quadratics-mcq04-4}
\end{center}
\end{figure}
~ ~ Explanation:
So close! You've got the general direction correct (the positive U-shape), though the $x$-intercepts are slightly off. You may have gotten confused with the many minus signs in the equation, so let's take a closer look together.
The quadratic seems to be in factorised form, though on further inspection the $(1 - x)$ factor looks slightly out of place. To properly read it in factorised form, we need to change $-x$ into $x$. How would we do that?
\begin{align*}
y &= -(x - 3)(1 - x) \newline
\Rightarrow y &= -(x - 3)[-(x - 1)]
\end{align*}
Looking at this, we only have to factorise a $-1$ out of that factor to transform $-x$ to $x$. By simplifying, we'll get the correct answer.
Just from this much working out, we can see that one of the factors is $(x - 3)$. Don't forget that the opposite sign rule applies here: we need to switch the sign in front of the number inside the factor to get the $x$-intercept. This means that one of the intercepts should be at $(3, 0)$, whereas this graph shows an intercept at $(-3, 0)$. Try to simplify the rest of the equation and apply this process to the second factor!
\par\null\par\null\par\null
\section*{Question 4 Hint Menu}\label{question-4-hint-menu}
Copy/paste the following tutorials:
\begin{itemize}
\tightlist
\item
\emph{~Quadratic functions in factorised form}
\item
\emph{Summary of quadratic functions in all their forms}
\end{itemize}
\textbf{Question analysis}
This question is asking us to find the graph that corresponds to the function we're given. Where do we start? Let's take a look at the function first, since this will tell us, in some way, what the graph will look like. We should identify what form it's given to us in: factorised form.
Even though it's got two factors, it looks slightly different to the factorised form itself. Specifically, one of the factors is $(1 - x)$, which doesn't look like it fits the form very well. We should manipulate the factors so that we can get the function perfectly in factorised form.
Once we've done that, we can have a look at what information the form and the equation tells us. Go back to the tutorials on this if you're not sure. Based on this information, which graph best matches what the function tells us?
\pagebreak
\section*{Question 5:~}\label{question-5}
\textbf{Title:} Calculate the range of a quadratic.
\textbf{Tags:}~range, quadratic
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
\textbf{Prompt:}
\par\null
The range of the quadratic defined as $\displaystyle f:(-\infty, 0] \to R, f(x) = \frac{1}{2}(x + 2)^2 - 1$ is:
\par\null
A:
\par\null
$[-2 - \sqrt{2}, -2 + \sqrt{2}]$
~ ~ Explanation:
This isn't quite correct.
To have reached this answer, you probably substituted $y = 0$ into the equation, or more specifically, made $f(x) = 0$, giving the following equation:
\[0 = \frac{1}{2}(x + 2)^2 - 1\]
Solving this would get the two solutions contained in this answer. Thinking about this, what does this solution actually mean? By subbing in $f(x) = 0$, we're actually finding $x$-intercepts rather than $y$ values. You may have wanted to calculate the $y$ value when $x = 0$, but accidentally found the $x$ values when $y = 0$! Be careful with what you sub in!
A method to fix this issue is to get in the habit of asking yourself ``what will I get if I sub this in?'' By consciously thinking of this, solutions will make more sense when we come across more methods for solving equations to find particular information.
To answer this question, we need to think about what the lowest and the highest $y$ values are. Once we have these we can construct an interval to represent our range. The problem is, how do we find these? The method that's usually the most efficient is to sketch out the graph, which we can do by hand since the function is helpfully in turning point form. Don't forget that turning point form is awesome when it comes to sketching since it tells us the general direction of the graph ($a > 0 \Rightarrow$ positive, U-shaped parabola) and the turning point of the graph.
Using the opposite sign rule inside the quadratic term, we can see that the turning point should be $(-2, -1)$. Plotting its graph should give us something as follows. Remember to stop the graph at $x = 0$ with a closed circle since the domain ends as an inclusive endpoint.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.56\columnwidth]{figures/quadratics-mc05-12/quadratics-mc05-12}
\caption{{\{\{s3\_url\}\}/mc-05/quadratics\_mc05\_1.png
{\label{683848}}%
}}
\end{center}
\end{figure}
Let's take a look at what this graph tells us --- specifically, we need to find the lowest and the highest points.
Firstly, even though the endpoint is found at $y = 1$, it isn't actually the lowest point. We can see the graph going further below the $x$-axis all the way down to the turning point. Finding the exact $y$ value of this turning point has something to do with the form that our function is given in.
Secondly, as $x$ goes to the left, the $y$ values seem to head further up. In other words, $y$ tends to $+\infty$. We can write this mathematically as well:
\[\text{As} \; x \to -\infty, \; y \to +\infty\]
Work out what these values are, put them into interval notation and you'll have your answer!
\par\null\par\null
B:
\par\null
$(-\infty, -1]$
~ ~ Explanation:
This isn't quite right! It's great that you've identified the turning point as being at $y=-1$, but from there it looks like you got a bit confused.
The best way, and sometimes the most efficient way, to work out the range is to see the graph of function. We can sketch this by hand since the function is helpfully in turning point form. Don't forget that turning point form is awesome when it comes to sketching since it tells us the general direction of the graph ($a > 0 \Rightarrow$ positive, U-shaped parabola) and the turning point of the graph.
Using the opposite sign rule inside the quadratic term, we can see that the turning point should be $(-2, -1)$. Plotting its graph should give us something as follows. Remember to stop the graph at $x = 0$ with a closed circle since the domain ends as an inclusive endpoint.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.56\columnwidth]{figures/quadratics-mc05-11/quadratics-mc05-11}
\caption{{\{\{s3\_url\}\}/mc-05/quadratics\_mc05\_1.png
{\label{883764}}%
}}
\end{center}
\end{figure}
The first thing to notice is the turning point --- which you've correctly identified already. The second thing to note is the direction which the $y$ values head towards. When the $x$ values head off towards the left, it seems as though the $y$ values head upwards toward positive $\infty$. We can also write this mathematically like this:
\[\text{As} \; x \to -\infty, \; y \to +\infty\]
Now we know what the lowest $y$ value is, as well as which way the $y$ values head toward. Let's use this to construct an answer that's hopefully one of the other three options!
\par\null\par\null
C:
\par\null
$[1, \infty)$
~ ~ Explanation:
That's really close, though it's not correct! You've done well to identify the endpoint, though the endpoint isn't always the lowest point in the range.
A \textbf{common misconception} is that the endpoint gives us the lowest point; however, this isn't always the case. We can quickly see if we're right or wrong just by looking a the graph. We can sketch this by hand since the function is helpfully in turning point form. Don't forget that turning point form is awesome when it comes to sketching since it tells us the general direction of the graph ($a > 0 \Rightarrow$ positive, U-shaped parabola) and the turning point of the graph.
Using the opposite sign rule inside the quadratic term, we can see that the turning point should be $(-2, -1)$. Plotting its graph should give us something as follows. Remember to stop the graph at $x = 0$ with a closed circle since the domain ends as an inclusive endpoint.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.56\columnwidth]{figures/quadratics-mc05-1/quadratics-mc05-1}
\caption{{\{\{s3\_url\}\}/mc-05/quadratics\_mc05\_1.png
{\label{729733}}%
}}
\end{center}
\end{figure}
Just by looking at the graph, we can see that the endpoint is actually on the part of the graph that comes back upwards. An even lower point than the endpoint exists, and we can see this at the turning point.
We can get the correct answer by using this turning point and its $y$ value, and combining it with which direction the $y$ values tend toward. Try using this to get the answer.
D: (correct)
\par\null
$[-1, \infty)$
~ ~ Explanation:
Yes, that's correct!
When calculating ranges based on restricted domains, it's best to have a look at its graph first --- this provides us with a clear picture of what the range should be. We can do this by hand since the function is helpfully in turning point form. Don't forget that turning point form is awesome when it comes to sketching since it tells us the general direction of the graph ($a > 0 \Rightarrow$ positive, U-shaped parabola) and the turning point of the graph.
Using the opposite sign rule inside the quadratic term, we can see that the turning point should be $(-2, -1)$. Plotting its graph should give us something as follows. Remember to stop the graph at $x = 0$ with a closed circle since the domain ends as an inclusive endpoint.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.56\columnwidth]{figures/quadratics-mcq05-1/quadratics-mc05-1}
\caption{{\{\{s3\_url\}\}/mc-05/quadratics\_mc05\_1.png
{\label{249876}}%
}}
\end{center}
\end{figure}
From the graph, we can see that even though the endpoint is has a $y = 1$, a lower point can be found at the turning point. From the function in turning point form, the lowest point is at $y = -1$.
Using the fact that the graph extends up to $+\infty$ as $x$ heads to the left, we can say that the range is $[-1, \infty)$
\par\null\par\null
\section*{Question 5 Hint Menu}\label{question-5-hint-menu}
Copy/paste the following tutorials:
\begin{itemize}
\tightlist
\item
\emph{~The definition of a quadratic}
\item
\emph{~Quadratic functions in turning point form}
\end{itemize}
\textbf{Question analysis}
\par\null
In this question, we're asked to calculate the range of a quadratic function. Remember that the range refers to the set of all the possible values that $y$ can take.
If this quadratic were on its maximal domain, this would be easy! All we'd have to do is go back to our definition of a quadratic and what its domain and range are. However, we're not given a maximal domain, rather we're given a restricted domain of $(-\infty, 0]$. What do we need to think of when doing this question?
First of all, if we can see \textbf{the graph of this function}, it would make it much easier to find the range --- all we'd have to do is look at how high or low the graph goes. One option in solving this question is doing a rough sketch of the graph on its domain. Consider \textbf{the form of this function} and what information it provides to help you quickly sketch the graph.
A second option is to do some calculations. What kind of calculations, though? Let's put it this way: \textbf{the domain affects the range}, since $y$ depends on $x$. If the domain is restricted, that means the graph must stop at an endpoint. If we calculate the $y$ value of this endpoint, then compare it with the theoretical lowest point on the graph (the turning point), we should be able to work out what the lowest point of the range is. Now, the question is how do we find the $y$ value of the endpoint, and how do we know what the lowest point is?
\begin{itemize}
\item The $x$ value of the endpoint gives us the $y$ value of the endpoint by substituting it into...
\item The lowest point of the range is the turning point of the graph. Looking at the form of this function, what can we say about the turning point?
\end{itemize}
\par\null
\par\null\par\null
\pagebreak
\section*{Question 6:~}\label{question-6}
\textbf{Title:} Convert a quadratic equation to turning point form.
\textbf{Tags:}~factorised form, turning point form, expand, completing
the square
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
\textbf{Prompt:}
\par\null\par\null
Converting the quadratic expression $\displaystyle -\frac{1}{8}(x + \frac{7}{2})(x - \frac{9}{2})$ to turning point form would give us:
\par\null
A: (correct)
\par\null
$\displaystyle -\frac{1}{8} \left( x - \frac{1}{2} \right) ^2 + 2$
~ ~ Explanation:
Yes, that's it! You're correct.
There are multiple ways of doing this, including:
\begin{itemize}
\item Converting from \textbf{factorised form} to \textbf{expanded form} by expanding out the factors. Then, completing the square to convert from \textbf{expanded form} to \textbf{turning point form}. This is arguably the quickest method, despite doing two stages of working out.
\item If you've realised before that mathematical operations can almost always be undone, then you might realise that if we can go from turning point form to factorised form by using the difference of two squares, we should be able to go the opposite way. This involves more thinking, however, as you have to undo the simplification steps of difference of two squares.
\end{itemize}
Since the first method is usually the quickest we'll touch on that one specifically. All we have to do is expand the two factors first.
\begin{align*}
-\frac{1}{8}(x + \frac{7}{2})(x - \frac{9}{2}) &= -\frac{1}{8}(x^2 - \frac{9}{2}x +\frac{7}{2}x - \frac{63}{4}) \newline
&= -\frac{1}{8}(x^2 - x -\frac{63}{4})
\end{align*}
Here's a \textbf{quick tip} when taking the next step. Don't expand the $\displaystyle -\frac{1}{8}$ into the rest of the expression. This is because you will have to factorise that back out when completing the square. Remember that to complete the square we must always factorise the coefficient of $x^2$ our before starting, and in this example it's conveniently set up for us!
The rest of the working out involves completing the square, remembering to always expand the $\displaystyle -\frac{1}{8}$ back out at the very end. This will give you this correct answer.
\par\null
B:
\par\null
$\displaystyle -\frac{1}{8} \left( x - \frac{1}{2} \right) ^2 - 16$
~ ~ Explanation:
\textbf{}
This is so close, though it's not correct! You've done everything correctly to get this answer, it's just the last step in your working out that let you down. Let's take a closer look.
\begin{align*}
&-\frac{1}{8}(x^2 - x - \frac{63}{4}) \qquad \qquad \quad \; \; \, \text{(This is the result from expanding)} \newline
&= -\frac{1}{8}[(x - \frac{1}{2})^2 - \frac{1}{4} - \frac{63}{4}] \newline
&= -\frac{1}{8}[(x - \frac{1}{2})^2 - 16] \qquad \qquad \; \, \text{(This is the result from completing the square)} \newline
\Rightarrow &\neq -\frac{1}{8}(x - \frac{1}{2})^2 - 16
\end{align*}
It's here that the error was most likely made: when the $\displaystyle -\frac{1}{8}$ was expanded back into the rest of the expression, the $16$ wasn't multiplied by the $\displaystyle -\frac{1}{8}$. That's why the constant at the end of the expression should be slightly different.
If you fix this up, you should get the correct answer!
C:
\par\null
$\displaystyle -\frac{1}{8} \left( x - \frac{1}{2} \right) ^2 + \frac{31}{16}$
~ ~ Explanation:
This isn't quite right. You applied the correct idea, this error may have been due to a mix up with your positive and negative signs while you were completing the square. Completing the square is arguably the most efficient method to do this question. You would have done everything right to complete the square, so let's check out where you slipped up:
\begin{align*}
&-\frac{1}{8}(x^2 - x - \frac{63}{4}) \qquad \qquad \quad \; \; \, \text{(This is the result from expanding)} \newline
&= -\frac{1}{8}[(x - \frac{1}{2})^2 - \frac{1}{4} - \frac{63}{4}] \newline
\Rightarrow &\neq -\frac{1}{8}[(x - \frac{1}{2})^2 - \frac{62}{4}]
\end{align*}
It's the last two lines that are the most important here. We had to subtract a number from a \textbf{negative} number, which makes the result more negative. Let's take a look at this calculation specifically:
\begin{align*}
&-\frac{1}{4} - \frac{63}{4} \newline
&= -(\frac{1}{4} + \frac{63}{4})
\end{align}
Since both numbers have negative signs in front, we can factorise a negative sign out of both of them. Once we do that, we can see that it's simpler to add the numbers together. All we have to do then is to put this constant back into our original expression and complete the working out!
\par\null
D:
\par\null
$\displaystyle -\frac{1}{8} \left( x - \frac{1}{2} \right) ^2 + \frac{127}{32}$
~ ~ Explanation:
\par\null
This is close, though it's not quite correct! There may have been an error in the expansion of the original expression. Let's take a look.
All parts of this solution is correct except for the constant $\displaystyle \frac{127}{32}$ at the end. It's likely that you expanded the question's expression, then looked to complete the square afterwards; this is arguably the most efficient method to do this question. Although you completed the square correctly, it's just the original expansion that's let down, so let's check out where you slipped up:
\begin{align*}
&-\frac{1}{8}(x + \frac{7}{2})(x - \frac{9}{2}) \newline
\Rightarrow &\neq -\frac{1}{8}(x^2 - x - \frac{63}{2})
\end{align*}
The issue here is the fractional constant inside the brackets. Don't forget that when you multiply two fractions together, you have to multiply the top with the top and the bottom with the bottom:
\[\frac{7}{2} \times -\frac{9}{2} = - \frac{7 \times 9}{2 \times 2}\]
If we do this step correctly, then complete the square as well, we should be able to get the right answer!
\par\null\par\null
\section*{Question 6 Hint Menu}\label{question-6-hint-menu}
Copy/paste the following tutorials:
\begin{itemize}
\tightlist
\item
\emph{~Quadratic functions in turning point form}
\item
\emph{Summary of quadratic functions in all their forms}
\end{itemize}
\textbf{Question analysis}
This question is asking us to convert from one quadratic form to another. In our tutorials, we've covered the methods to switch between the three forms. If you take a look at the diagram, however, you will notice there's no direct way to go from factorised form to turning point form. This doesn't mean that it's impossible, though.
How do we approach this question? For starters, think about what we \textit{can} do with the factorised form and start from there. Can we convert from factorised to something else, then convert that to turning point form? Think outside of what's simply told by the conversion diagram from the tutorial, and you might find some tricks of your own!
\par\null\par\null\par\null
\pagebreak
\section*{Question 7:~}\label{question-7}
\textbf{Title:~}Calculate/identify/deduce
\textbf{Tags:}
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
A:
~ ~ Explanation:
\par\null
B:
~ ~ Explanation:
\par\null
C:
~ ~ Explanation:
\par\null
D:
~ ~ Explanation:
\par\null
\pagebreak
\section*{Question 8:~}\label{question-8}
\textbf{Title:~}Calculate/identify/deduce
\textbf{Tags:}
\textbf{Multiple choice options:~}(write `correct' in brackets after the
correct option)
A:
~ ~ Explanation:
\par\null
B:
~ ~ Explanation:
\par\null
C:
~ ~ Explanation:
\par\null
D:
~ ~ Explanation:
\par\null
~
\pagebreak
\subsection*{Question 1}\label{question-1}
\textbf{Summary Title:~}Calculate and work with the components of a
quadratic equation.
\textbf{Tag}:~quadratic equations, unknowns, sketching
\textbf{Source}: Connect
\textbf{Context:~}
\textbf{}
Consider the the function $f$, where $\displaystyle f: R \to R, f(x) = \frac{2}{3}x^2 + bx - 10$ and $b \in R$.
\par\null
\textbf{Question Part (a):}
\textbf{Prompt:~}
\par\null
The graph of $y = f(x)$ passes through the point $(4, 6)$.
Find the value of $b$.
\par\null
\textbf{Display type:~}Short answer-box
\textbf{No. marks:~}1
\par\null
\textbf{Question Part (b):}
\textbf{Prompt:~}~
\par\null
Hence, find the coordinates of the turning point on the graph of $y = f(x)$.
\par\null
\textbf{Display type:~}Short answer-box
\textbf{No. marks:} 3
\par\null
\textbf{Question Part (c):}
\textbf{Prompt:~}~
\par\null
Solve the equation $f(x) = 0$ for $x$.
\par\null
\textbf{Display type:~}Short answer-box
\textbf{No. marks:} 2
\par\null
~\textbf{Question Part (d):}
\textbf{Prompt:~}~
\par\null
Sketch the graph of $y = f(x)$, labelling any stationary points and axial intercepts in coordinate form.
\par\null
\textbf{Display type:~}Short answer-box
\textbf{No. marks:} 4
\par\null
\textbf{Question Part (a)}
\textbf{Marking step 1}
\textbf{Description of marks:~ ~ ~}Find the constant's value.
{[}1 mark for correctly substituting the coordinate to calculate the
unknown constant{]}
\textbf{~\textbf{Option type: Single}}
~\textbf{Marking prompt}~: Did you substitute the coordinate into the
equation and isolate the unknown constant to get b=4/3?
~\textbf{Options:~}
\begin{itemize}
\tightlist
\item
yes
\item
no
\end{itemize}
\textbf{Worked solution:}
\par\null
\begin{align*}
&Let \; y = f(x) \newline
&Sub \; (4, 6) \newline
\Rightarrow 6 &= \frac{2}{3}(4)^2 + b(4) - 10 \newline
6 + 10 &= \frac{2}{3}(16) + 4b \newline
16 &= \frac{32}{3} + 4b \newline
16 - \frac{32}{3} &= 4b \newline
\frac{48}{3} - \frac{32}{3} &= 4b \newline
\frac{16}{3} &= 4b \newline
\therefore b &= \frac{4}{3}
\end{align*}
\textbf{Question Part (b)}
\textbf{Marking step 1}
\textbf{Description of marks:~ ~ ~}Find the coordinates of the turning
point
{[}1 mark for using a method to find the x value of the turning point{]}
{[}1 mark for finding the y value of the turning point{]}
{[}1 mark for expressing the answer in coordinate form{]}
\textbf{~\textbf{Option type:~}}Multiple (checkboxes)~
~\textbf{Marking prompt}~: Which of the following calculations did you
perform?
~\textbf{Options:~}
\par\null
\begin{itemize}
\item I used the formula to calculate the $x$ value of the turning point ($x = -1$), OR I converted the function to turning point form to work out the $x$ value of the turning point.
\item I subbed in my $x$ value to find the $y$ value of the turning point ($\displaystyle y = -\frac{32}{3}$), OR I converted the function to turning point form to work out the $y$ value of the turning point.
\item I wrote my answer in coordinate form $\displaystyle (-1, -\frac{32}{3})$
\end{itemize}
\par\null
\textbf{Worked solution:}
\textbf{}
\begin{align*}
x &= -\frac{b}{2a} \newline
&where \; a = \frac{2}{3} \; and \; b = \frac{4}{3} \newline
\Rightarrow x &= -\frac{\frac{4}{3}}{2(\frac{2}{3})} \newline
x &= -\frac{\; \frac{4}{3} \;}{\; \frac{4}{3} \;} \newline
\Rightarrow x &= -1 \newline
Sub \; x &= -1 \; into \; y = f(x) \newline
\Rightarrow y &= \frac{2}{3}(-1)^2 + \frac{4}{3}(-1) - 10 \newline
y &= \frac{2}{3} - \frac{4}{3} - 10 \newline
y &= -\frac{2}{3} - 10 \newline
y &= -\frac{2}{3} - \frac{30}{3} \newline
\Rightarrow y &= -\frac{32}{3} \newline
\newline
&\therefore \; \text{turning point at} \; (-1, -\frac{32}{3})
\end{align*}
\textbf{Question Part (c)}
\textbf{Marking step 1}
\textbf{Description of marks:~ ~ ~}Find solutions for x
{[}1 mark for using any appropriate method to solve a quadratic
equation{]}
{[}1 mark for the two solutions{]}
\textbf{~\textbf{Option type:~}}Multiple (checkboxes)~
~\textbf{Marking prompt}~: Which of the following calculations did you
perform?
~\textbf{Options:~}
\textbf{}
\begin{itemize}
\item I used the cross method, completing the square or the quadratic formula to solve the equation, or any other method to solve quadratic equations.
\item I got $x = -5 \; or \; 3$ as my answer.
\end{itemize}
\par\null
\textbf{Worked solution:}
\textbf{}
\begin{align*}
\frac{2}{3}x^2 + \frac{4}{3}x - 10 &= 0 \newline
\frac{2}{3}(x^2 + 2x - 15) &= 0 \newline
x^2 + 2x - 15 &= 0 \newline
(x + 5)(x - 3) &= 0 \newline
\Rightarrow x + 5 &= 0 \; or \; x - 3 = 0 \newline
\therefore x &= -5 \; or \; 3
\end{align*}
\textbf{Question Part (d)}
\textbf{Marking step 1}
\textbf{Description of marks:~ ~ ~}Sketch the graph
{[}1 mark for sketching the graph with the correct shape: positive,
U-shaped parabola and intersects the graph twice{]}
{[}1 mark for labeling the stationary point in coordinate form{]}
{[}1 mark for labeling both x-intercepts in coordinate form{]}
{[}1 mark for labeling the y-intercept in coordinate form{]}
\textbf{~\textbf{Option type:~}}Multiple (checkboxes)~
~\textbf{Marking prompt}~: Which of the following calculations did you
perform?
~\textbf{Options:~}
\textbf{}
\begin{itemize}
\item I sketched the quadratic as a positive, U-shaped parabola that intersects the $x$-axis twice.
\item I labelled the turning point of the parabola with the coordinates $(-1, -\frac{32}{3})$.
\item I labelled the two $x$-intercepts in coordinate form: one with $(-5, 0)$, which was on the left of the $y$-axis; and the other with $(3, 0)$, which was on the right of the $y$-axis.
\item I labelled the $y$-intercept with the coordinates $(0, -10)$, which was underneath the $x$-axis.
\end{itemize}
\par\null
\textbf{Worked solution:}
\textbf{}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/quadratics-sa01-graph1/quadratics-sa01-graph1}
\caption{{\{\{s3\_url\}\}/sa01/quadratics\_sa01\_graph1.png
{\label{886638}}%
}}
\end{center}
\end{figure}
\section*{Question 1 Hint Menu}\label{question-1-hint-menu}
Copy/paste the following tutorials:
\begin{itemize}
\tightlist
\item
\emph{Quadratic functions in factorised form.}
\item
\emph{Quadratic functions in turning point form.~}
\item
\emph{Components of the quadratic formula}
\end{itemize}
\textbf{Question analysis}
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This set of questions involve using a quadratic function's equation to find its key characteristics.
\textbf{For part (a):} We can solve unknown constants if we have a coordinate which the graph goes through. This can be used by subbing in the $x$ and $y$ values then isolating the constant.
\textbf{For part (b):} We associate multiple things with the turning point, though to find its coordinates we have a couple of options. We have a formula derived from the quadratic formula that allows us to find the $x$ value of the turning point straight away; we can then use that $x$ value to find the $y$ value by simply subbing it back into the function. Otherwise, our other option to find the coordinate is to transform the entire function into turning point form, which gives us the turning point automatically!
\textbf{In regards to part (c):} solving quadratic equations can be done in a variety of ways, though it's best to pick and use a method that you're most comfortable with, whether that's the cross method, completing the square or the quadratic formula. If you're unsure about the different ways of solving these equations, check back to Topic: Algebraic techniques to brush up on your skills.
\textbf{Finally, for part (d):} when it comes to sketching graphs, we need to keep a few things in mind. Firstly, one way to approach a sketch is to \textbf{draw out the general shape first}, THEN fill in the coordinates of stationary points and intercepts. So based on this method we need to get the shape correct: this includes whether the parabola is positive or negative, how many intercepts it has and where these intercepts are located. Once you've got a general idea of what the graph will look like, sketch it out. Usually previous questions have gotten you to work out key characteristics, such as axial intercepts and turning points, so with this in hand we can then label in all the things the question has asked for.
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