# 0.6 Quadratic functions and graphs  (Page 2/2)

 Page 2 / 2
$\begin{array}{ccc}\hfill g\left(x\right)& =& {\left(x-1\right)}^{2}+2\hfill \\ \hfill {y}_{int}& =& {\left(0-1\right)}^{2}+2\hfill \\ & =& {\left(-1\right)}^{2}+2\hfill \\ & =& 1+2\hfill \\ & =& 3\hfill \end{array}$

The $x$ -intercepts are calculated as follows:

$\begin{array}{ccc}\hfill y& =& a{\left(x+p\right)}^{2}+q\hfill \\ \hfill 0& =& a{\left({x}_{int}+p\right)}^{2}+q\hfill \\ \hfill a{\left({x}_{int}+p\right)}^{2}& =& -q\hfill \\ \hfill {x}_{int}+p& =& ±\sqrt{-\frac{q}{a}}\hfill \\ \hfill {x}_{int}& =& ±\sqrt{-\frac{q}{a}}-p\hfill \end{array}$

However, [link] is only valid if $-\frac{q}{a}>0$ which means that either $q<0$ or $a<0$ but not both. This is consistent with what we expect, since if $q>0$ and $a>0$ then $-\frac{q}{a}$ is negative and in this case the graph lies above the $x$ -axis and therefore does not intersect the $x$ -axis. If however, $q>0$ and $a<0$ , then $-\frac{q}{a}$ is positive and the graph is hat shaped with turning point above the $x$ -axis and should have two $x$ -intercepts. Similarly, if $q<0$ and $a>0$ then $-\frac{q}{a}$ is also positive, and the graph should intersect with the $x$ -axis twice.

For example, the $x$ -intercepts of $g\left(x\right)={\left(x-1\right)}^{2}+2$ are given by setting $y=0$ to get:

$\begin{array}{ccc}\hfill g\left(x\right)& =& {\left(x-1\right)}^{2}+2\hfill \\ \hfill 0& =& {\left({x}_{int}-1\right)}^{2}+2\hfill \\ \hfill -2& =& {\left({x}_{int}-1\right)}^{2}\hfill \end{array}$

which has no real solutions. Therefore, the graph of $g\left(x\right)={\left(x-1\right)}^{2}+2$ does not have any $x$ -intercepts.

## Intercepts

1. Find the x- and y-intercepts of the function $f\left(x\right)={\left(x-4\right)}^{2}-1$ .
2. Find the intercepts with both axes of the graph of $f\left(x\right)={x}^{2}-6x+8$ .
3. Given: $f\left(x\right)=-{x}^{2}+4x-3$ . Calculate the x- and y-intercepts of the graph of $f$ .

## Turning points

The turning point of the function of the form $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is given by examining the range of the function. We know that if $a>0$ then the range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left[q,\infty \right)\right\}$ and if $a<0$ then the range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left(-\infty ,q\right]\right\}$ .

So, if $a>0$ , then the lowest value that $f\left(x\right)$ can take on is $q$ . Solving for the value of $x$ at which $f\left(x\right)=q$ gives:

$\begin{array}{ccc}\hfill q& =& a{\left(x+p\right)}^{2}+q\hfill \\ \hfill 0& =& a{\left(x+p\right)}^{2}\hfill \\ \hfill 0& =& {\left(x+p\right)}^{2}\hfill \\ \hfill 0& =& x+p\hfill \\ \hfill x& =& -p\hfill \end{array}$

$\therefore$ $x=-p$ at $f\left(x\right)=q$ . The co-ordinates of the (minimal) turning point is therefore $\left(-p,q\right)$ .

Similarly, if $a<0$ , then the highest value that $f\left(x\right)$ can take on is $q$ and the co-ordinates of the (maximal) turning point is $\left(-p,q\right)$ .

## Turning points

1. Determine the turning point of the graph of $f\left(x\right)={x}^{2}-6x+8$ .
2. Given: $f\left(x\right)=-{x}^{2}+4x-3$ . Calculate the co-ordinates of the turning point of $f$ .
3. Find the turning point of the following function by completing the square: $y=\frac{1}{2}{\left(x+2\right)}^{2}-1$ .

## Axes of symmetry

There is only one axis of symmetry for the function of the form $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ . This axis passes through the turning point and is parallel to the $y$ -axis. Since the $x$ -coordinate of the turning point is $x=-p$ , this is the axis of symmetry.

## Axes of symmetry

1. Find the equation of the axis of symmetry of the graph $y=2{x}^{2}-5x-18$ .
2. Write down the equation of the axis of symmetry of the graph of $y=3{\left(x-2\right)}^{2}+1$ .
3. Write down an example of an equation of a parabola where the y-axis is the axis of symmetry.

## Sketching graphs of the form $f\left(x\right)=a{\left(x+p\right)}^{2}+q$

In order to sketch graphs of the form $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ , we need to determine five characteristics:

1. sign of $a$
2. domain and range
3. turning point
4. $y$ -intercept
5. $x$ -intercept

For example, sketch the graph of $g\left(x\right)=-\frac{1}{2}{\left(x+1\right)}^{2}-3$ . Mark the intercepts, turning point and axis of symmetry.

Firstly, we determine that $a<0$ . This means that the graph will have a maximal turning point.

The domain of the graph is $\left\{x:x\in \mathbb{R}\right\}$ because $f\left(x\right)$ is defined for all $x\in \mathbb{R}$ . The range of the graph is determined as follows:

$\begin{array}{ccc}\hfill {\left(x+1\right)}^{2}& \ge & 0\hfill \\ \hfill -\frac{1}{2}{\left(x+1\right)}^{2}& \le & 0\hfill \\ \hfill -\frac{1}{2}{\left(x+1\right)}^{2}-3& \le & -3\hfill \\ \hfill \therefore f\left(x\right)& \le & -3\hfill \end{array}$

Therefore the range of the graph is $\left\{f\left(x\right):f\left(x\right)\in \left(-\infty ,-3\right]\right\}$ .

Using the fact that the maximum value that $f\left(x\right)$ achieves is -3, then the $y$ -coordinate of the turning point is -3. The $x$ -coordinate is determined as follows:

$\begin{array}{ccc}\hfill -\frac{1}{2}{\left(x+1\right)}^{2}-3& =& -3\hfill \\ \hfill -\frac{1}{2}{\left(x+1\right)}^{2}-3+3& =& 0\hfill \\ \hfill -\frac{1}{2}{\left(x+1\right)}^{2}& =& 0\hfill \\ \hfill \mathrm{Divide both sides by}\phantom{\rule{2pt}{0ex}}-\frac{1}{2}:\phantom{\rule{1.em}{0ex}}{\left(\mathrm{x}+1\right)}^{2}& =& 0\hfill \\ \hfill \mathrm{Take square root of both sides:}\phantom{\rule{1.em}{0ex}}\mathrm{x}+1& =& 0\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}x& =& -1\hfill \end{array}$

The coordinates of the turning point are: $\left(-1,-3\right)$ .

The $y$ -intercept is obtained by setting $x=0$ . This gives:

$\begin{array}{ccc}\hfill {y}_{int}& =& -\frac{1}{2}{\left(0+1\right)}^{2}-3\hfill \\ & =& -\frac{1}{2}\left(1\right)-3\hfill \\ & =& -\frac{1}{2}-3\hfill \\ & =& -\frac{1}{2}-3\hfill \\ & =& -\frac{7}{2}\hfill \end{array}$

The $x$ -intercept is obtained by setting $y=0$ . This gives:

$\begin{array}{ccc}\hfill 0& =& -\frac{1}{2}{\left({x}_{int}+1\right)}^{2}-3\hfill \\ \hfill 3& =& -\frac{1}{2}{\left({x}_{int}+1\right)}^{2}\hfill \\ \hfill -3·2& =& {\left({x}_{int}+1\right)}^{2}\hfill \\ \hfill -6& =& {\left({x}_{int}+1\right)}^{2}\hfill \end{array}$

which has no real solutions. Therefore, there are no $x$ -intercepts.

We also know that the axis of symmetry is parallel to the $y$ -axis and passes through the turning point. Graph of the function f ( x ) = - 1 2 ( x + 1 ) 2 - 3

## Sketching the parabola

1. Draw the graph of $y=3{\left(x-2\right)}^{2}+1$ showing all the intercepts with the axes as well as the coordinates of the turning point.
2. Draw a neat sketch graph of the function defined by $y=a{x}^{2}+bx+c$ if $a>0$ ; $b<0$ ; ${b}^{2}=4ac$ .

## Writing an equation of a shifted parabola

Given a parabola with equation $y={x}^{2}-2x-3$ . The graph of the parabola is shifted one unit to the right. Or else the y-axis shifts one unit to the left i.e. $x$ becomes $x-1$ . Therefore the new equation will become:

$\begin{array}{ccc}\hfill y& =& {\left(x-1\right)}^{2}-2\left(x-1\right)-3\hfill \\ & =& {x}^{2}-2x+1-2x+2-3\hfill \\ & =& {x}^{2}-4x\hfill \end{array}$

If the given parabola is shifted 3 units down i.e. $y$ becomes $y+3$ . The new equation will be:

(Notice the x-axis then moves 3 units upwards)

$\begin{array}{ccc}\hfill y+3& =& {x}^{2}-2x-3\hfill \\ \hfill y& =& {x}^{2}-2x-6\hfill \end{array}$

## End of chapter exercises

1. Show that if $a<0$ , then the range of $f\left(x\right)=a{\left(x+p\right)}^{2}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left(-\infty ,q\right]\right\}$ .
2. If (2,7) is the turning point of $f\left(x\right)=-2{x}^{2}-4ax+k$ , find the values of the constants $a$ and $k$ .
3. The graph in the figure is represented by the equation $f\left(x\right)=a{x}^{2}+bx$ . The coordinates of the turning point are (3,9). Show that $a=-1$ and $b=6$ .
4. Given: $y={x}^{2}-2x+3$ . Give the equation of the new graph originating if:
1. The graph of $f$ is moved three units to the left.
2. The $x$ -axis is moved down three units.
5. A parabola with turning point (-1,-4) is shifted vertically by 4 units upwards. What are the coordinates of the turning point of the shifted parabola?

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