0.6 Quadratic functions and graphs  (Page 2/2)

The $x$ -intercepts are calculated as follows:

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

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

Intercepts

1. Find the x- and y-intercepts of the function $f\left(x\right)={(x-4)}^2-1$ .
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 $f\left(x\right)=a{(x+p)}^2+q$ is given by examining the range of the function. We know that if $a>0$ then the range of $f\left(x\right)=a{(x+p)}^2+q$ is $\left\{f\right(x):f(x)\in [q,\infty \left)\right\}$ and if $a<0$ then the range of $f\left(x\right)=a{(x+p)}^2+q$ is $\left\{f\right(x):f(x)\in (-\infty ,q\left]\right\}$ .

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:

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

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 $(-p,q)$ .

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: $y=\frac{1}{2}{(x+2)}^2-1$ .

Axes of symmetry

There is only one axis of symmetry for the function of the form $f\left(x\right)=a{(x+p)}^2+q$ . 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 $y=2x^2-5x-18$ .
2. Write down the equation of the axis of symmetry of the graph of $y=3{(x-2)}^2+1$ .
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 $f\left(x\right)=a{(x+p)}^2+q$

In order to sketch graphs of the form $f\left(x\right)=a{(x+p)}^2+q$ , 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 $g\left(x\right)=-\frac{1}{2}{(x+1)}^2-3$ . 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 $\{x:x\in \mathbb{R}\}$ because $f\left(x\right)$ is defined for all $x\in \mathbb{R}$ . The range of the graph is determined as follows:

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

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:

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

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

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

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.

Sketching the parabola

1. Draw the graph of $y=3{(x-2)}^2+1$ 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:

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)

End of chapter exercises

1. Show that if $a<0$ , then the range of $f\left(x\right)=a{(x+p)}^2+q$ is $\left\{f\right(x):f(x)\in (-\infty ,q\left]\right\}$ .
2. If (2,7) is the turning point of $f\left(x\right)=-2x^2-4ax+k$ , 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?