Linear and Quadratic Functions

Lessons

Senior 3 · Mathematics

Linear and Quadratic Functions

5 lessons

Lesson 4 of 5

Quadratic functions: tables of values, vertex and axis of symmetry

25 min

The shape of a flight

Photograph a bouncing ball with a flash that fires 25 times a second, and each flight between bounces traces an arch. The arch is symmetric, and the ball crowds together at the top, where it moves most slowly. The curve is (very nearly) a parabola, the graph of a quadratic function.

In Unit 2 you solved quadratic equations. Now you study the quadratic function as a whole: its table of values, its highest or lowest point, and its line of symmetry.

Definition

Quadratic function: a function of the form

f(x) = ax^2 + bx + c, \qquad a \ne 0,

where $a$ , $b$ and $c$ are real numbers. Its graph is a curve called a parabola.

Example: $f (x) = x^{2} - 2 x - 3$ has $a = 1$ , $b = - 2$ , $c = - 3$ . The function $g (x) = 5 x - 1$ is not quadratic: it has no $x^{2}$ term.

A table of values

To draw $y = x^{2} - 2 x - 3$ , work out $y$ for whole-number values of $x$ . One row per term keeps the arithmetic safe.

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

Plot the points and join them with a smooth curve, not straight segments. The $y$ -values repeat in mirror order about $x = 1$ : 5, 0, $- 3$ , then $- 4$ , then $- 3$ , 0, 5.

The graph of y = x² − 2x − 3 from its table of values.

Definition

Vertex of a parabola: its turning point, the lowest point if $a > 0$ or the highest point if $a < 0$ . Its $y$ -coordinate is the minimum (or maximum) value of the function.

Axis of symmetry: the vertical line through the vertex. The parabola is its own mirror image in this line.

Example: $y = x^{2} - 2 x - 3$ has vertex $(1, - 4)$ , minimum value $- 4$ , and axis of symmetry $x = 1$ .

Important

Direction and vertex of $f (x) = a x^{2} + b x + c$

If $a > 0$ the parabola opens upward ( $\cup$ ) and has a minimum; if $a < 0$ it opens downward ( $\cap$ ) and has a maximum.
The axis of symmetry and the vertex are

x = -\frac{b}{2a}, \qquad \text{vertex } \left( -\frac{b}{2a},\; f\!\left(-\frac{b}{2a}\right) \right).

Why: complete the square, as in Unit 2:

ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}.

The square is never negative and equals 0 only when $x = -\frac{b}{2a}$ . So if $a > 0$ , $f$ is smallest there; if $a < 0$ , it is largest there. Moving the same distance to the left or right of $-\frac{b}{2a}$ gives the same square, so the graph is symmetric about that line.

In vertex form, $f (x) = a (x - h)^{2} + k$ , the vertex is $(h, k)$ and the axis is $x = h$ .

Worked example 1: vertex by the formula, checked by completing the square

Find the vertex, the axis of symmetry and the minimum value of $f (x) = 2 x^{2} - 8 x + 5$ .

Solution $a = 2 > 0$ , so the parabola opens upward and has a minimum.

x = -\frac{b}{2a} = -\frac{-8}{2 \times 2} = 2, \qquad f(2) = 2(4) - 16 + 5 = -3.

Vertex $(2, - 3)$ , axis $x = 2$ , minimum value $- 3$ .

Check: $2x^2 - 8x + 5 = 2(x^2 - 4x) + 5 = 2\big[(x - 2)^2 - 4\big] + 5 = 2(x - 2)^2 - 3$ ✓.

Worked example 2: a parabola that opens downward

For $g (x) = - x^{2} + 4 x + 1$ , find the vertex and axis, make a table of values for $-1 \le x \le 5$ , and draw the graph.

Solution $a = - 1 < 0$ : it opens downward. $x = -\frac{4}{2(-1)} = 2$ and $g (2) = - 4 + 8 + 1 = 5$ . Vertex $(2, 5)$ , a maximum; axis $x = 2$ .

$x$	$- 1$	$0$	$1$	$2$	$3$	$4$	$5$
$g (x)$	$- 4$	1	4	5	4	1	$- 4$

The values are symmetric about x = 2, and the vertex (2, 5) is the highest point.

Worked example 3: the highest point of a throw

A ball is thrown straight up. Its height after $t$ seconds is modelled by $h (t) = 20 t - 5 t^{2}$ metres. Find the greatest height and when it lands.

Solution $a = - 5 < 0$ , so there is a maximum at $t = -\frac{20}{2(-5)} = 2$ s, where $h (2) = 40 - 20 = 20$ m.

It lands when $h = 0$ : $5 t (4 - t) = 0$ , so $t = 0$ (the throw) or $t = 4$ s. The top is at $t = 2$ , exactly halfway between the two roots: the axis of symmetry lies midway between the $x$ -intercepts.

Worked example 4: the largest pen

A farmer has 40 m of fencing to make a rectangular pen for goats against a straight wall, so only three sides need fencing. Find the dimensions that give the largest area.

Solution Let each side at right angles to the wall be $x$ m. The side parallel to the wall is $40 - 2 x$ m, so

A(x) = x(40 - 2x) = -2x^2 + 40x, \qquad 0 < x < 20.

$a = - 2 < 0$ , so there is a maximum at $x = -\frac{40}{2(-2)} = 10$ : $A(10) = 10 \times 20 = 200\ \mathrm{m^2}$ .

The pen should be 10 m deep and 20 m along the wall, with area $200\ \mathrm{m^2}$ .

Warning

Three slips with the vertex formula

Forgetting the minus: for $2 x^{2} - 8 x + 5$ , $-\frac{b}{2a} = -\frac{-8}{4} = +2$ , not $- 2$ .
Forgetting the 2: it is $2 a$ underneath, not $a$ .
Stopping halfway: $x = 2$ is only the axis. Substitute to get the $y$ -coordinate of the vertex.

And read the sign of $a$ first: when $a < 0$ the vertex gives a maximum.

Tip

Use the symmetry to halve the work. Once you know the axis $x = h$ , every value you compute at $h + d$ is also the value at $h - d$ . A table centred on the vertex needs only half the arithmetic.

Khan Academy: Parabola vertex and axis of symmetry. Watch how completing the square shows where the vertex is, and why the parabola is symmetric about the vertical line through it.

✦

Practice

The last three are harder.

Copy and complete the table of values of $y = x^{2} - 4 x + 3$ for $x = - 1, 0, 1, 2, 3, 4, 5$ , and draw the graph.
Say whether each parabola opens upward or downward: (a) $y = 3 x^{2} - x$ (b) $y = 5 + 2 x - x^{2}$ .
Find the vertex and the axis of symmetry of $y = x^{2} + 6 x + 5$ .
Find the maximum value of $f (x) = - 2 x^{2} + 12 x - 7$ , and the value of $x$ where it occurs.
Write the vertex of (a) $y = (x - 3)^{2} + 2$ (b) $y = - 2 (x + 1)^{2} - 4$ .
A parabola has $x$ -intercepts $- 3$ and 5. What is its axis of symmetry?
A stone's height is $h (t) = 15 t - 5 t^{2}$ metres. Find its greatest height.
With 60 m of fencing a farmer encloses a rectangle (all four sides). Show that the area is $A = x (30 - x)$ , and find the greatest possible area.

Show answersHide answers

$y$ : 8, 3, 0, $- 1$ , 0, 3, 8. Vertex $(2, - 1)$ .
(a) upward, $a = 3 > 0$ (b) downward, $a = - 1 < 0$ .
$x = -\frac{6}{2} = -3$ ; $y = 9 - 18 + 5 = - 4$ . Vertex $(- 3, - 4)$ , axis $x = - 3$ .
$x = -\frac{12}{2(-2)} = 3$ ; $f (3) = - 18 + 36 - 7 = 11$ . Maximum 11 at $x = 3$ .
(a) $(3, 2)$ (b) $(- 1, - 4)$ .
Midway between the roots: $x = \frac{-3 + 5}{2} = 1$ .
$t = -\frac{15}{2(-5)} = 1.5$ s; $h = 22.5 - 11.25 = 11.25$ m.
If one side is $x$ , then $2 x + 2 y = 60$ gives $y = 30 - x$ , so $A = x (30 - x) = - x^{2} + 30 x$ . Maximum at $x = 15$ : $A = 225\ \mathrm{m^2}$ , a square.

Sources and credits

Written by Ganzaa.
Photo: Bouncing ball strobe edit.jpg, MichaelMaggs, edited by Richard Bartz, Wikimedia Commons, licensed under CC BY-SA 3.0. Used unchanged.
Video: Parabola vertex and axis of symmetry, Khan Academy, embedded from YouTube.

The tables, graphs, worked examples, practice items and quiz questions were written by Ganzaa; the throw and pen problems use illustrative figures. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1The shape of a flight

2A table of values

The shape of a flight

A table of values