Graphs of Quadratic Functions

What a quadratic function $y = ax^2 + bx + c$ is, how to make its table of values and draw its graph to a given scale, and how the sign of $a$ decides whether the curve opens upwards or downwards.

Before you start

In chapter 2 you drew straight-line graphs from a table of values: work out $y$ for each value of $x$ , plot the points to the scale given, and join them.
Substituting a negative number needs brackets: when $x = -2$ , $x^2 = (-2)^2 = 4$ and $-3x = -3 \times (-2) = 6\text{.}$

Quadratic Functions

In $y = x^2 - 2x - 3$ , each value of $x$ gives exactly one value of $y$ : when $x = 4$ , $y = 16 - 8 - 3 = 5$ . We say that $y$ is a function of $x$ . Because the highest power of $x$ is $2$ , it is a quadratic function.

A quadratic function

A quadratic function is a function $y = ax^2 + bx + c$ , where $a$ , $b$ and $c$ are numbers and $a \neq 0$ .

Its graph is a smooth, symmetrical curve called a parabola.

Example 1: Naming $a$ , $b$ and $c$

State the values of $a$ , $b$ and $c$ for
(a) $y = 2x^2 - 3$
(b) $y = x^2 + 4x$
(c) $y = 3 + 2x - x^2$ .

Solution

(a) There is no $x$ term, so $b = 0$ : $a = 2$ , $b = 0$ , $c = -3$ .

(b) There is no constant term: $a = 1$ , $b = 4$ , $c = 0$ .

(a) $a = 2$ , $b = 0$ , $c = -3$
(b) $a = 1$ , $b = 4$ , $c = 0$
(c) $a = -1$ , $b = 2$ , $c = 3$

Tables of Values

A question gives the range of $x$ , such as $-2 \le x \le 4$ . Work out $y$ for every whole number in that range. It helps to work out each term in its own row, and then add the rows.

Example 2: A table, term by term

Complete a table of values for $y = x^2 - 2x - 3$ for $-2 \le x \le 4$ .

Solution

For each $x$ , work out $x^2$ , then $-2x$ , then add them and $-3$ . For example, at $x = -2$ : $x^2 = (-2)^2 = 4$ , $-2x = -2 \times (-2) = 4\text{,}$ so $y = 4 + 4 - 3 = 5$ .

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$x^2$	$4$	$1$	$0$	$1$	$4$	$9$	$16$
$-2x$	$4$	$2$	$0$	$-2$	$-4$	$-6$	$-8$
$-3$	$-3$	$-3$	$-3$	$-3$	$-3$	$-3$	$-3$
$y$	$5$	$0$	$-3$	$-4$	$-3$	$0$	$5$

Check one value straight from the equation: at $x = 3$ , $y = 9 - 6 - 3 = 0$ ✓.

The values of $y$ are $5, 0, -3, -4, -3, 0, 5\text{.}$

Example 3: A number in front of $x^2$

Complete a table of values for $y = 2x^2 - 3x - 2$ for $-2 \le x \le 3$ .

Solution

Square first, then multiply by $2$ : at $x = -2$ , $2x^2 = 2 \times 4 = 8$ (not $(2 \times (-2))^2 = 16$ ), and $-3x = 6$ . So $y = 8 + 6 - 2 = 12$ .

At $x = -1$ : $2 + 3 - 2 = 3$ . At $x = 0$ : $-2$ . At $x = 1$ : $2 - 3 - 2 = -3$ . At $x = 2$ : $8 - 6 - 2 = 0$ . At $x = 3$ : $18 - 9 - 2 = 7$ .

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$12$	$3$	$-2$	$-3$	$0$	$7$

The values of $y$ are $12, 3, -2, -3, 0, 7$ .

Slips to watch for

The square of a negative number is positive: $(-2)^2 = 4$ , not $-4$ .

In $-x^2$ , square first and then take the negative: at $x = -3$ , $-x^2 = -9$ , not $9$ .

A minus sign times a negative $x$ gives a positive term: at $x = -2$ , $-3x = +6$ .

In $2x^2$ , square first and then multiply: at $x = -2$ , $2x^2 = 8$ , not $16$ .

Drawing the Graph

Drawing the graph of a quadratic function

1. Draw and number the axes to the scale the question gives (for example, 2 cm to 1 unit on the $x$ -axis and 1 cm to 1 unit on the $y$ -axis), so that they cover the range of $x$ and every value of $y$ in the table.

2. Plot every point of the table.

3. Join the points with one smooth curve, drawn freehand through every point. Do not join them with straight lines, and round the bottom (or top) of the curve; it is not a sharp point.

4. Label the curve with its equation.

Example 4: Drawing the graph

Draw the graph of $y = x^2 - 2x - 3$ for $-2 \le x \le 4$ .

Solution

Use the table of Example 2: $(-2, 5)$ , $(-1, 0)$ , $(0, -3)$ , $(1, -4)$ , $(2, -3)$ , $(3, 0)$ , $(4, 5)$ . Here each square is 1 unit on both axes; on graph paper, use the scale the question gives.

The curve passes through every point and is ∪-shaped: it falls, turns at its lowest point, and rises again. The two halves are mirror images of each other. Lesson 5.6 finds the lowest point and the line of symmetry.

The graph is the smooth ∪-shaped curve above, drawn from $x = -2$ to $x = 4$ .

Opening Upwards or Downwards

In $y = x^2 - 2x - 3$ the coefficient of $x^2$ is positive, and the curve opens upwards. Now look at a function whose coefficient of $x^2$ is negative.

Example 5: A negative coefficient of $x^2$

Draw the graph of $y = 6 + x - x^2$ for $-3 \le x \le 4$ .

Solution

At $x = -3$ : $6 + (-3) - (-3)^2 = 6 - 3 - 9 = -6\text{.}$ At $x = -2$ : $6 - 2 - 4 = 0$ . At $x = 4$ : $6 + 4 - 16 = -6$ . The others in the same way:

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$-6$	$0$	$4$	$6$	$6$	$4$	$0$	$-6$

The curve is ∩-shaped: it rises, turns at its highest point, and falls. The highest point lies between $x = 0$ and $x = 1$ , where the table has two equal values, $6$ ; the curve goes a little above $6$ there, so do not draw it flat.

The graph is the smooth ∩-shaped curve above.

The shape of the graph of y = ax² + bx + c

If $a > 0$ , the curve is ∪-shaped (it opens upwards) and has a lowest point, its minimum point.

If $a < 0$ , the curve is ∩-shaped (it opens downwards) and has a highest point, its maximum point.

Find $a$ wherever the $x^2$ term is written: $y = 6 + x - x^2$ has $a = -1$ , so its graph is ∩-shaped.

The size of $a$ changes the width: the graph of $y = 2x^2$ is narrower than that of $y = x^2$ , and the graph of $y = \frac{1}{2}x^2$ is wider.

Where the Curve Cuts the y-Axis

On the $y$ -axis, $x = 0$ . Then $y = a(0)^2 + b(0) + c = c\text{.}$ So the graph of $y = ax^2 + bx + c$ cuts the $y$ -axis at $(0, c)$ , and $c$ is called the $y$ -intercept. In Example 4 the curve cuts the $y$ -axis at $(0, -3)$ ; in Example 5, at $(0, 6)$ .

The sign of $a$ and the value of $c$ together are often enough to tell graphs apart.

Example 6: Matching graphs to equations

The curves $A$ , $B$ , $C$ and $D$ are the graphs of these four equations, in some order:
(1) $y = x^2 + 2x - 3$
(2) $y = -x^2 + 2x + 3$
(3) $y = x^2 - 2x + 3$
(4) $y = -x^2 - 2x$ . Match each curve to its equation.

Solution

Equations (1) and (3) have $a = 1 > 0$ , so their graphs are ∪-shaped: $B$ and $D$ . Equations (2) and (4) have $a = -1 < 0$ : $A$ and $C$ .

Now the $y$ -intercepts. $B$ cuts the $y$ -axis at $-3$ , so $B$ is (1), with $c = -3$ ; $D$ cuts it at $3$ , so $D$ is (3).

$A$ cuts the $y$ -axis at $3$ , so $A$ is (2); $C$ passes through the origin, so $C$ is (4), with $c = 0$ .

Check one point: (2) at $x = 3$ gives $-9 + 6 + 3 = 0$ , and $A$ does cross the $x$ -axis at $3$ ✓.

$A$ is (2), $B$ is (1), $C$ is (4) and $D$ is (3).

Finding an Unknown Coefficient

A point lies on a curve exactly when its coordinates make the equation true. So if you know a point on the curve, you can substitute it to find a missing coefficient.

Example 7: A point on the curve

The curve $y = x^2 + bx - 3$ passes through the point $(2, 5)$ . Find $b$ .

Solution

Substitute $x = 2$ and $y = 5$ : $5 = 2^2 + 2b - 3$ , so $5 = 1 + 2b$ .

Subtract $1$ : $2b = 4$ , so $b = 2$ .

Check: $y = x^2 + 2x - 3$ at $x = 2$ gives $4 + 4 - 3 = 5$ ✓

$b = 2$ , so the curve is $y = x^2 + 2x - 3$ .

Common Mistakes

Joining the points with straight lines. A quadratic graph is one smooth curve.
Drawing the bottom (or top) of the curve as a sharp point, or flat between two equal values. It is rounded, and it may go beyond the table’s values, as in Example 5.
Squaring a negative number wrongly: $(-2)^2 = 4$ . And in $-x^2$ at $x = -3$ , the value is $-9$ .
Plotting a point with the wrong sign, such as $(-2, -5)$ for $(-2, 5)$ . A point far off a smooth curve is a sign of a slip: check its value.
Counting squares instead of reading the numbers on an axis whose scale is not 1 unit to a square.
Drawing the curve beyond the range of $x$ given, or stopping short of it.

Practice Questions

Use graph paper for the questions that ask you to draw. The questions get harder as you go.

Question 1

For each function, state $a$ , $b$ and $c$ , say whether its graph is ∪-shaped or ∩-shaped, and give the point where it cuts the $y$ -axis.
(a) $y = x^2 - 5x + 4$
(b) $y = 3 - 2x^2$
(c) $y = -x^2 + x$
(d) $y = 7 + x^2$

ShowHide solution

(a) $a = 1$ , $b = -5$ , $c = 4$ . $a > 0$ : ∪-shaped. It cuts the $y$ -axis at $(0, 4)$ .

(b) $y = -2x^2 + 3$ : $a = -2$ , $b = 0$ , $c = 3$ . $a < 0$ : ∩-shaped. $(0, 3)$ .

(d) $y = x^2 + 7$ : $a = 1$ , $b = 0$ , $c = 7$ . ∪-shaped. $(0, 7)$ .

(a) ∪, $(0, 4)$
(b) ∩, $(0, 3)$
(c) ∩, $(0, 0)$
(d) ∪, $(0, 7)$

Question 2

Copy and complete the table for $y = x^2 - 4x + 1$ , and draw the graph for $-1 \le x \le 5$ .

$x$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y$	$6$			$-3$			$6$

ShowHide solution

At $x = 0$ : $1$ . At $x = 1$ : $1 - 4 + 1 = -2$ . At $x = 3$ : $9 - 12 + 1 = -2$ . At $x = 4$ : $16 - 16 + 1 = 1$ .

$x$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y$	$6$	$1$	$-2$	$-3$	$-2$	$1$	$6$

The missing values are $1, -2, -2, 1$ ; the graph is the ∪-shaped curve above.

Question 3

Make a table of values for $y = 4x - x^2$ for $-1 \le x \le 5$ , and draw the graph. Is it ∪-shaped or ∩-shaped, and why?

ShowHide solution

At $x = -1$ : $-4 - 1 = -5$ . At $x = 0$ : $0$ . At $x = 1$ : $4 - 1 = 3$ . At $x = 2$ : $8 - 4 = 4$ . At $x = 3$ : $12 - 9 = 3$ . At $x = 4$ : $0$ . At $x = 5$ : $20 - 25 = -5$ .

$x$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y$	$-5$	$0$	$3$	$4$	$3$	$0$	$-5$

It is ∩-shaped, because the coefficient of $x^2$ is $-1$ , which is negative.

$y$ : $-5, 0, 3, 4, 3, 0, -5\text{;}$ ∩-shaped, since $a = -1 < 0$ .

Question 4

Make a table of values for $y = 2x^2 + x - 6$ for $-3 \le x \le 2$ , and draw the graph, with each square on the $y$ -axis standing for 2 units.

ShowHide solution

At $x = -3$ : $2 \times 9 - 3 - 6 = 9$ . At $x = -2$ : $8 - 2 - 6 = 0$ . At $x = -1$ : $2 - 1 - 6 = -5$ . At $x = 0$ : $-6$ . At $x = 1$ : $2 + 1 - 6 = -3$ . At $x = 2$ : $8 + 2 - 6 = 4$ .

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$
$y$	$9$	$0$	$-5$	$-6$	$-3$	$4$

The point $(-3, 9)$ lies halfway between the grid lines for $8$ and $10$ , and $(-1, -5)$ halfway between $-4$ and $-6$ . The lowest point of the curve is a little below $-6$ , between $x = -1$ and $x = 0$ .

$y$ : $9, 0, -5, -6, -3, 4\text{;}$ the graph is the ∪-shaped curve above.

Question 5

Ahmad worked out $y = -x^2 - 3x + 4$ at $x = -2$ like this: “ $-x^2 = 4$ and $-3x = -6$ , so $y = 4 - 6 + 4 = 2$ .” Find his two mistakes and the correct value of $y$ .

ShowHide solution

First mistake: $-x^2$ means the negative of $x^2$ . $x^2 = (-2)^2 = 4$ , so $-x^2 = -4$ , not $4$ .

Second mistake: $-3x = -3 \times (-2) = +6\text{,}$ not $-6$ .

So $y = -4 + 6 + 4 = 6$ .

$y = 6$

Question 6

Priya drew the graph of $y = x^2 - 2x - 3$ for $-2 \le x \le 4$ like this. Give three things that are wrong.

ShowHide solution

At $x = -2$ , $y = 4 + 4 - 3 = 5$ . She plotted $(-2, -5)$ instead of $(-2, 5)$ .

She joined the points with straight lines. The graph should be one smooth curve through the points.

Her graph has a sharp point at $(1, -4)$ . The curve should be rounded at its lowest point.

She has also not labelled the graph with its equation. Any three of these four faults is a full answer.

Compare it with the correct graph in Example 4.

Any three of: the point $(-2, 5)$ is plotted as $(-2, -5)$ ; the points are joined by straight lines; the bottom is a sharp point instead of a rounded curve; the graph is not labelled with its equation.

Question 7

The curve $y = x^2 - 3x + c$ passes through $(4, 1)$ . Find $c$ , and write down where the curve cuts the $y$ -axis.

ShowHide solution

Substitute: $1 = 16 - 12 + c$ , so $1 = 4 + c$ , and $c = -3$ .

The curve cuts the $y$ -axis at $(0, c) = (0, -3)$ .

Check: $16 - 12 - 3 = 1$ ✓

$c = -3$ ; the curve cuts the $y$ -axis at $(0, -3)$ .

Question 8

The curve $y = 2x^2 + bx + 1$ passes through $(-1, 5)$ . Find $b$ .

ShowHide solution

Substitute: $5 = 2(-1)^2 + b(-1) + 1 = 2 - b + 1 = 3 - b\text{.}$

So $b = 3 - 5 = -2$ .

Check: $y = 2x^2 - 2x + 1$ at $x = -1$ gives $2 + 2 + 1 = 5$ ✓

$b = -2$

Question 9

A ball is thrown straight up. Its height, $h$ metres, after $t$ seconds is given by $h = 20t - 5t^2$ . Make a table of values for $t = 0, 1, 2, 3, 4$ and draw the graph of $h$ against $t$ , with 5 m to a square on the $h$ -axis.

ShowHide solution

At $t = 1$ : $20 - 5 = 15$ . At $t = 2$ : $40 - 20 = 20$ . At $t = 3$ : $60 - 45 = 15$ . At $t = 4$ : $80 - 80 = 0$ .

$t$	$0$	$1$	$2$	$3$	$4$
$h$	$0$	$15$	$20$	$15$	$0$

The coefficient of $t^2$ is $-5$ , so the curve is ∩-shaped: the ball rises, then falls back to the ground at $t = 4$ .

$h$ : $0, 15, 20, 15, 0$ ; the graph is the ∩-shaped curve above.

Question 10

A rectangle has sides $x$ cm and $(6 - x)$ cm. Its area, $A$ cm², is $A = 6x - x^2$ . Make a table of values for $x = 0, 1, 2, 3, 4, 5, 6$ and draw the graph of $A$ against $x$ .

ShowHide solution

At $x = 1$ : $6 - 1 = 5$ . At $x = 2$ : $12 - 4 = 8$ . At $x = 3$ : $18 - 9 = 9$ . At $x = 4$ : $24 - 16 = 8$ . At $x = 5$ : $30 - 25 = 5$ . At $x = 0$ and $x = 6$ : $0$ .

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$A$	$0$	$5$	$8$	$9$	$8$	$5$	$0$

The curve is ∩-shaped (the coefficient of $x^2$ is $-1$ ). Only values of $x$ between $0$ and $6$ make sense: both sides, $x$ and $6 - x$ , must be more than 0 cm. The end points, where the area is $0$ , complete the curve.

$A$ : $0, 5, 8, 9, 8, 5, 0$ ; the graph is the ∩-shaped curve above.

Question 11

Without drawing, match each function to its description. Functions:
(a) $y = x^2 - 4$
(b) $y = 4 - x^2$
(c) $y = x^2 + 4x$
(d) $y = -x^2 + 4x - 5$ .
Descriptions:
(1) ∩-shaped, cuts the $y$ -axis at $(0, -5)$
(2) ∪-shaped, passes through the origin
(3) ∩-shaped, cuts the $y$ -axis at $(0, 4)$
(4) ∪-shaped, cuts the $y$ -axis at $(0, -4)$ .

ShowHide solution

(a) $a = 1 > 0$ , $c = -4$ : ∪-shaped through $(0, -4)$ , which is (4).

(b) $a = -1 < 0$ , $c = 4$ : ∩-shaped through $(0, 4)$ , which is (3).

(d) $a = -1 < 0$ , $c = -5$ : ∩-shaped through $(0, -5)$ , which is (1).

(a) and (4);
(b) and (3);
(c) and (2);
(d) and (1)

Key Points

A quadratic function is $y = ax^2 + bx + c$ with $a \neq 0$ ; its graph is a smooth curve called a parabola.
Make a table for every whole number in the range of $x$ given; take care with negative values of $x$ .
Draw the axes to the scale given, plot every point, and join them with one smooth curve; label it.
If $a > 0$ the graph is ∪-shaped with a minimum point; if $a < 0$ it is ∩-shaped with a maximum point.
The graph cuts the $y$ -axis at $(0, c)$ .
A point lies on the curve when its coordinates make the equation true; use this to find a missing coefficient.

Before you start

Quadratic Functions

A quadratic function

Example 1: Naming aaa, bbb and ccc

Tables of Values

Example 2: A table, term by term

Example 3: A number in front of x2x^2x2

Slips to watch for

Drawing the Graph

Drawing the graph of a quadratic function

Example 4: Drawing the graph

Opening Upwards or Downwards

Example 5: A negative coefficient of x2x^2x2

The shape of the graph of y = ax² + bx + c

Where the Curve Cuts the y-Axis

Example 6: Matching graphs to equations

Finding an Unknown Coefficient

Example 7: A point on the curve

Common Mistakes

Practice Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Key Points

Example 1: Naming $a$ , $b$ and $c$

Example 3: A number in front of $x^2$

Example 5: A negative coefficient of $x^2$