Home Practice
For learners and parents For teachers and schools
Textbooks
Full catalogue
Leaderboards
Learners Leaderboard Classes/Grades Leaderboard Schools Leaderboard
Pricing Support
Help centre Contact us
Log in

We think you are located in United States. Is this correct?

6.3 Quadratic functions

6.3 Quadratic functions (EMA4H)

Functions of the form \(y={x}^{2}\) (EMA4J)

Functions of the general form \(y=a{x}^{2}+q\) are called parabolic functions. In the equation \(y=a{x}^{2}+q\), \(a\) and \(q\) are constants and have different effects on the parabola.

Worked example 4: Plotting a quadratic function

\(y=f(x)={x}^{2}\)

Complete the following table for \(f(x)={x}^{2}\) and plot the points on a system of axes.

\(x\)

\(-\text{3}\)

\(-\text{2}\)

\(-\text{1}\)

\(\text{0}\)

\(\text{1}\)

\(\text{2}\)

\(\text{3}\)

\(f(x)\)

\(\text{9}\)

  1. Join the points with a smooth curve.

  2. The domain of \(f\) is \(x\in \mathbb{R}\). Determine the range.

  3. About which line is \(f\) symmetrical?

  4. Determine the value of \(x\) for which \(f(x)=6\frac{1}{4}\). Confirm your answer graphically.

  5. Where does the graph cut the axes?

Substitute values into the equation

\begin{align*} f(x) & = {x}^{2} \\ f(-3) & = {(-3)}^{2} = 9\\ f(-2) & = {(-2)}^{2} = 4\\ f(-1) & = {(-1)}^{2} = 1\\ f(0) & = {(0)}^{2} = 0\\ f(1) & = {(1)}^{2} = 1\\ f(2) & = {(2)}^{2} = 4\\ f(3) & = {(3)}^{2} = 9 \end{align*}

\(x\)

\(-\text{3}\)

\(-\text{2}\)

\(-\text{1}\)

\(\text{0}\)

\(\text{1}\)

\(\text{2}\)

\(\text{3}\)

\(f(x)\)

\(\text{9}\)

\(\text{4}\)

\(\text{1}\)

\(\text{0}\)

\(\text{0}\)

\(\text{4}\)

\(\text{9}\)

Plot the points and join with a smooth curve

From the table, we get the following points:

\((-3;9), (-2;4), (-1;1), (0;0), (1;1), (2;4), (3;9)\)
39d0806f7b3ca29607019d0deea1c686.png

Determine the domain and range

Domain: \(x\in \mathbb{R}\)

From the graph we see that for all values of \(x\), \(y \ge 0\).

Range: \(\left\{y:y\in \mathbb{R}, y\ge 0\right\}\)

Find the axis of symmetry

\(f\) is symmetrical about the \(y\)-axis. Therefore the axis of symmetry of \(f\) is the line \(x=0\).

Determine the \(x\)-value for which \(f(x)=6\frac{1}{4}\)

\begin{align*} f(x)& = \frac{25}{4} \\ \therefore \frac{25}{4}& = {x}^{2} \\ x& = ±\frac{5}{2}\\ & = ±2\frac{1}{2} \end{align*}

See points \(A\) and \(B\) on the graph.

Determine the intercept

The function \(f\) intercepts the axes at the origin \((0;0)\).

We notice that as the value of \(x\) increases from \(-\infty\) to \(\text{0}\), \(f(x)\) decreases.

At the turning point \((0;0)\), \(f(x)=0\).

As the value of \(x\) increases from \(\text{0}\) to \(∞\), \(f(x)\) increases.

Functions of the form \(y=a{x}^{2}+q\) (EMA4K)

The effects of \(a\)and \(q\) on a parabola.

Complete the table and plot the following graphs on the same system of axes:

  1. \({y}_{1}={x}^{2}-2\)

  2. \({y}_{2}={x}^{2}-1\)

  3. \({y}_{3}={x}^{2}\)

  4. \({y}_{4}={x}^{2}+1\)

  5. \({y}_{5}={x}^{2}+2\)

\(x\)

\(-\text{2}\)

\(-\text{1}\)

\(\text{0}\)

\(\text{1}\)

\(\text{2}\)

\({y}_{1}\)

\({y}_{2}\)

\({y}_{3}\)

\({y}_{4}\)

\({y}_{5}\)

Use your results to deduce the effect of \(q\).

Complete the table and plot the following graphs on the same system of axes:

  1. \({y}_{6}=-2{x}^{2}\)

  2. \({y}_{7}=-{x}^{2}\)

  3. \({y}_{8}={x}^{2}\)

  4. \({y}_{9}=2{x}^{2}\)

\(x\)

\(-\text{2}\)

\(-\text{1}\)

\(\text{0}\)

\(\text{1}\)

\(\text{2}\)

\({y}_{6}\)

\({y}_{7}\)

\({y}_{8}\)

\({y}_{9}\)

Use your results to deduce the effect of \(a\).

The effect of \(q\)

The effect of \(q\) is called a vertical shift because all points are moved the same distance in the same direction (it slides the entire graph up or down).

  • For \(q>0\), the graph of \(f(x)\) is shifted vertically upwards by \(q\) units. The turning point of \(f(x)\) is above the \(y\)-axis.

  • For \(q<0\), the graph of \(f(x)\) is shifted vertically downwards by \(q\) units. The turning point of \(f(x)\) is below the \(y\)-axis.

The effect of \(a\)

The sign of \(a\) determines the shape of the graph.

  • For \(a>0\), the graph of \(f(x)\) is a “smile” and has a minimum turning point at \((0;q)\). The graph of \(f(x)\) is stretched vertically upwards; as \(a\) gets larger, the graph gets narrower.

    For \(0<a<1\), as \(a\) gets closer to \(\text{0}\), the graph of \(f(x)\) gets wider.

  • For \(a<0\), the graph of \(f(x)\) is a “frown” and has a maximum turning point at \((0;q)\). The graph of \(f(x)\) is stretched vertically downwards; as \(a\) gets smaller, the graph gets narrower.

    For \(-1<a<0\), as \(a\) gets closer to \(\text{0}\), the graph of \(f(x)\) gets wider.

fbce0cdcff7c7ffaf43990f1fbe4dd57.png

\(a<0\)

\(a>0\)

\(q>0\)

17c7dccddff80b26b12c6db7ad5251f5.png 7a6047096fb79bd9c3f6e419cd234545.png

\(q=0\)

4515d28ba5e6c4df8f7f4c78b787fc49.png 983d0ac25a8482b4daf6dd2a4a60bdb2.png

\(q<0\)

f200a5305fdf7451039da68839b788a0.png 7d43fef3d5b2913285c2eee6779627f3.png
Table 6.2: The effect of \(a\) and \(q\) on a parabola.

You can use this Phet simulation to help you see the effects of changing \(a\) and \(q\) for a parabola.

Discovering the characteristics (EMA4M)

The standard form of the equation of a parabola is \(y=a{x}^{2}+q\).

Domain and range

The domain is \(\left\{x:x\in \mathbb{R}\right\}\) because there is no value for which \(f(x)\) is undefined.

If \(a>0\) then we have:

\[\begin{array}{cccl} {x}^{2}& \ge & 0 & (\text{Perfect square is always positive}) \\ a{x}^{2}& \ge & 0 & (\text{since } a>0) \\ a{x}^{2}+q & \ge & q & (\text{add } q \text{ to both sides}) \\ \therefore f(x) & \ge & q & \end{array}\]

Therefore if \(a>0\), the range is \(\left[q;\infty \right)\). Similarly, if \(a<0\) then the range is \(\left(-\infty ;q\right]\).

Worked example 5: Domain and range of a parabola

If \(g(x)={x}^{2}+2\), determine the domain and range of the function.

Determine the domain

The domain is \(\left\{x:x\in \mathbb{R}\right\}\) because there is no value for which \(g(x)\) is undefined.

Determine the range

The range of \(g(x)\) can be calculated as follows:

\begin{align*} {x}^{2}& \ge 0 \\ {x}^{2}+2& \ge 2\\ g(x)& \ge 2 \end{align*}

Therefore the range is \(\left\{g\left(x\right):g\left(x\right)\ge 2\right\}\).

Intercepts

The \(y\)-intercept:

Every point on the \(y\)-axis has an \(x\)-coordinate of \(\text{0}\), therefore to calculate the \(y\)-intercept let \(x=0\).

For example, the \(y\)-intercept of \(g(x)={x}^{2}+2\) is given by setting \(x=0\):

\begin{align*} g(x)& = {x}^{2}+2 \\ g(0)& ={0}^{2}+2\\ & =2 \end{align*}

This gives the point \((0;2)\).

The \(x\)-intercepts:

Every point on the \(x\)-axis has a \(y\)-coordinate of \(\text{0}\), therefore to calculate the \(x\)-intercept let \(y=0\).

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

\begin{align*} g(x)& = {x}^{2}+2 \\ 0 & = {x}^{2}+2 \\ -2& = {x}^{2} \end{align*}

There is no real solution, therefore the graph of \(g(x)={x}^{2}+2\) does not have \(x\)-intercepts.

Turning points

The turning point of the function of the form \(f(x)=a{x}^{2}+q\) is determined by examining the range of the function.

  • If \(a>0\), the graph of \(f(x)\) is a “smile” and has a minimum turning point at \((0;q)\).

  • If \(a<0\), the graph of \(f(x)\) is a “frown” and has a maximum turning point at \((0;q)\).

Axes of symmetry

The axis of symmetry for functions of the form \(f(x)=a{x}^{2}+q\) is the \(y\)-axis, which is the line \(x=0\).

Sketching graphs of the form \(y=a{x}^{2}+q\) (EMA4N)

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

  1. sign of \(a\)

  2. \(y\)-intercept

  3. \(x\)-intercept

  4. turning point

Worked example 6: Sketching a parabola

Sketch the graph of \(y={2x}^{2}-4\). Mark the intercepts and turning point.

Examine the standard form of the equation

We notice that \(a>0\). Therefore the graph is a “smile” and has a minimum turning point.

Calculate the intercepts

For the \(y\)-intercept, let \(x=0\):

\begin{align*} y& = 2{x}^{2}-4 \\ & = 2{(0)}^{2}-4 \\ & =-4 \end{align*}

This gives the point \((0;-4)\).

For the \(x\)-intercepts, let \(y=0\):

\begin{align*} y& = 2{x}^{2}-4 \\ 0& = 2{x}^{2}-4 \\ {x}^{2}& = 2 \\ \therefore x& = ±\sqrt{2} \end{align*}

This gives the points \((-\sqrt{2};0)\) and \((\sqrt{2};0)\).

Determine the turning point

From the standard form of the equation we see that the turning point is \((0;-4)\).

Plot the points and sketch the graph

67e1f7292a65ec49c1b85a25c2479edb.png

Domain: \(\left\{x:x\in \mathbb{R}\right\}\)

Range: \(\left\{y:y\ge -4,y\in \mathbb{R}\right\}\)

The axis of symmetry is the line \(x=0\).

Worked example 7: Sketching a parabola

Sketch the graph of \(g(x)=-\frac{1}{2}{x}^{2}-3\). Mark the intercepts and the turning point.

Examine the standard form of the equation

We notice that \(a<0\). Therefore the graph is a “frown” and has a maximum turning point.

Calculate the intercepts

For the \(y\)-intercept, let \(x=0\):

\begin{align*} g(x)& = -\frac{1}{2}{x}^{2}-3 \\ g(0)& = -\frac{1}{2}{(0)}^{2}-3 \\ & = -3 \end{align*}

This gives the point \((0;-3)\).

For the \(x\)-intercepts let \(y=0\):

\begin{align*} 0& = -\frac{1}{2}{x}^{2}-3 \\ 3& = -\frac{1}{2}{x}^{2} \\ -2(3)& = {x}^{2} \\ -6& = {x}^{2} \end{align*}

There is no real solution, therefore there are no \(x\)-intercepts.

Determine the turning point

From the standard form of the equation we see that the turning point is \((0;-3)\).

Plot the points and sketch the graph

0c79830b28e6f1a7ca6ef728c186fa26.png

Domain: \(x\in \mathbb{R}\).

Range: \(y\in \left(-\infty ;-3\right]\).

The axis of symmetry is the line \(x=0\).

Textbook Exercise 6.3

The graph below shows a quadratic function with the following form: \(y = ax^2 + q\). Two points on the parabola are shown: Point A, the turning point of the parabola, at \((0;4)\), and Point B is at \(\left(2; \frac{8}{3}\right)\). Calculate the values of \(a\) and \(q\).

b3949e7e56dc64e38ca3d7cd27e77921.png

The value of \(q\) is 4.

\begin{align*} y & = ax^2 +4 \\ \left( \frac{8}{3} \right) & = a(\text{2})^2 +4 \quad \longleftarrow \quad ^{\text{substitute in the}}_{\text{coordinates of a point!}} \\ \frac{8}{3} & = \text{4}a +4 \\ \frac{8}{3} -4 & = \text{4}a \\ - \frac{4}{3} & = \text{4}a \\ - \frac{1}{3} & = a \end{align*} \[a = - \frac{1}{3};~ q = 4\]

The graph below shows a quadratic function with the following form: \(y = ax^2 + q\). Two points on the parabola are shown: Point A, the turning point of the parabola, at \((0;-3)\), and Point B is at \(\left(2; 5\right)\). Calculate the values of \(a\) and \(q\).

248c7d2b1605f5654d751cc3dac0bbf2.png

The value of \(q\) is -3.

\begin{align*} y & = ax^2 -3 \\ \left( 5 \right) & = a(\text{2})^2 -3 \quad \longleftarrow \quad ^{\text{substitute in the}}_{\text{coordinates of a point!}} \\ 5 & = \text{4}a -3 \\ 5 +3 & = \text{4}a \\ 8 & = \text{4}a \\ 2 & = a \end{align*} \[a = 2;~ q = -3\]

Given the following equation:

\(y=5 x^{2} - 2\)

Calculate the \(y\)-coordinate of the \(y\)-intercept.

\begin{align*} y & = ax^2 + q \\ & = 5 x^{2} - 2 \\ & = 5 (0)^{2} - 2\\ & = 0-2 \end{align*}

The \(y\)-coordinate of the \(y\)-intercept is \(-2\).

Now calculate the \(x\)-intercepts. Your answer must be correct to 2 decimal places.

\begin{align*} y & = 5 x^{2} - 2 \\ (0) & =5 x^{2} - 2 \\ - 5 x^{2} &=-2\\ x^2 &= \frac{-2}{-5} \\ x & =\pm \sqrt{\frac{2}{5}}\\ \text{Therefore: } x = +\sqrt{\frac{2}{5}} &\text{ and } x = - \sqrt{\frac{2}{5}} \\ x= -\text{0,63} &\text{ and } x= \text{0,63} \end{align*}

The \(x\)-intercepts are \((-\text{0,63};0)\) and \((\text{0,63};0)\).

Given the following equation:

\(y=- 2 x^{2} + 1\)

Calculate the \(y\)-coordinate of the \(y\)-intercept.

\begin{align*} y & = ax^2 + q \\ & = - 2 x^{2} + 1 \\ & = - 2 (0)^{2} + 1\\ & = 0 + 1 \end{align*}

The \(y\)-coordinate of the \(y\)-intercept is \(\text{1}\).

Now calculate the \(x\)-intercepts. Your answer must be correct to 2 decimal places.

\begin{align*} y & = - 2 x^{2} + 1 \\ (0) & =- 2 x^{2} + 1 \\ 2 x^{2} &=1\\ x^2 &= \frac{1}{2} \\ x & =\pm \sqrt{\frac{1}{2}}\\ \text{Therefore: } x = +\sqrt{\frac{1}{2}} &\text{ and } x = - \sqrt{\frac{1}{2}} \\ x= -\text{0,71} & \text{ and } x\end{align*}

The \(x\)-intercepts are \((-\text{0,71};0)\) and \((\text{0,71};0)\).

Given the following graph, identify a function that matches each of the following equations:

8182ca8a7041c9d8be12625042598dab.png

\(y = \text{0,5}x^2\)

\(h(x)\)

\(y = x^2\)

\(g(x)\)

\(y = 3x^2\)

\(f(x)\)

\(y = -x^2\)

\(k(x)\)

Given the following graph, identify a function that matches each of the following equations:

d7c530cac9d4f26ef9bb9085a0e93407.png

\(y = x^2 - 3\)

\(h(x)\)

\(y = x^2 + 1\)

\(f(x)\)

\(y = x^2\)

\(g(x)\)

Two parabolas are drawn: \(g: y=ax^2+p\) and \(h:y=bx^2+q\).

cb51eb52e1618caa0332c02a39a81f67.png

Find the values of \(a\) and \(p\).

\(p\) is the \(y\)-intercept of the function \(g(x)\), therefore \(p=-9\)

To find \(a\) we use one of the points on the graph (e.g. \((4;7)\)):

\begin{align*} y&=ax^2-9\\ 7&= a(4^2) - 9\\ 16a&=16\\ \therefore a&=1 \end{align*}

\(a=1\); \(p=-9\)

Find the values of \(b\) and \(q\).

\(q\) is the \(y\)-intercept of the function \(h(x)\), therefore \(q = 23\)

To find \(b\), we use one of the points on the graph (e.g. \((4;7)\)):

\begin{align*} y&=bx^{2} =23\\ 7&= b(4^{2}) +23\\ 16b&=-16\\ \therefore b&=-1 \end{align*}

\(b=-1\); \(q=23\)

Find the values of \(x\) for which \(g(x) \geq h(x)\).

These are the points where \(g\) lies above \(h\).

From the graph we see that \(g\) lies above \(h\) when: \(x \le -4\) or \(x \geq 4\)

For what values of \(x\) is \(g\) increasing?

\(g\) increases from the turning point \((0;-9)\), i.e. for \(x \geq 0\).

Show that if \(a < 0\) the range of \(f(x)=ax^2 + q\) is \(\left\{f(x):f(x) \le q\right\}\).

Because the square of any number is always positive we get: \(x^2 \geq 0\).

If we multiply by \(a\) where \((a < 0)\) then the sign of the inequality is reversed: \(ax^2 \le 0\)

Adding \(q\) to both sides gives \(ax^2 + q \le q\)

And so \(f(x) \le q\)

This gives the range as \((-\infty; q]\).

Draw the graph of the function \(y=-x^2 + 4\) showing all intercepts with the axes.

The \(y\)-intercept is \((0;4)\). The \(x\)-intercepts are given by setting \(y = 0\):

\begin{align*} 0 & = -x^2 + 4 \\ x^2 & = 4 \\ x & = \pm 2 \end{align*}

Therefore the \(x\)-intercepts are: \((2;0)\) and \((-2;0)\).

Now we can sketch the graph:

a338e005090157dbcde8dc2082d13b85.png