Parabolas: vertex, roots and the quadratic formula

You kick a ball high across the field. It rises, slows down, turns and falls back down. The path it makes is called a parabola. You see the same shape in a jet of water from a garden hose.

Say the ball is y metres high when it is x metres away, with y = −0.05x² + x. By the end of this post you will work out how high the ball goes and where it lands. First, see how a formula like this works.

Three letters: a, b and c

You can always write a quadratic function like this:

y = ax² + bx + c

Its graph is a parabola. Move the sliders for a, b and c below. The graph starts as y = x² − 6x + 1.

Move a, b and c

a = 1b = −6c = 1

y = x² − 6x + 1

• Graph
• Axis of symmetry
• Vertex
• Roots
• y-intercept

Shape

opens upward

Axis of symmetry

x = −b ÷ 2a = 3

Vertex

(3, −8)

y-intercept

(0, 1)

Discriminant

D = b² − 4ac = 32

Roots (2)

x ≈ 0.17 and x ≈ 5.83

Show the numbers

x	y
−1	8
0	1
1	−4
2	−7
3	−8
4	−7
5	−4
6	1
7	8

This is what each letter does:

• a sets the shape. If a is positive, the parabola opens upward and has a lowest point. If a is negative, it opens downward and has a highest point. The further a is from 0, the narrower the parabola.
• With a = 0 the x² drops out. You are left with y = bx + c: a straight line.
• c moves the graph up or down. Put in x = 0 and only c is left. So the graph always crosses the y-axis at (0, c).
• b is the tricky one. Change only b and the vertex slides off at an angle. The vertex itself then moves along a parabola: y = −ax² + c. That is the same shape, upside down.

Finding the vertex

Every parabola is symmetric. Its left side is the mirror image of its right side. The axis of symmetry is the vertical line through the vertex.

Same height, same distance

• y = x² − 6x + 1
• Axis of symmetry x = 3
• Points at the same height
• Vertex

Show the numbers

x	y
0	1
1	−4
2	−7
3	−8
4	−7
5	−4
6	1

Look at (0, 1) and (6, 1). They are at the same height, so the axis lies exactly halfway between them: at x = 3. You do not have to hunt for points. A formula gives you the axis too:

x = −b ÷ 2a

Take y = x² − 6x + 1. Here a = 1, b = −6 and c = 1.

x = −(−6) ÷ (2 × 1) = 6 ÷ 2 = 3
y = 3² − 6 × 3 + 1 = 9 − 18 + 1 = −8

The vertex is (3, −8). Because a is positive, it is the lowest point.

Second way: completing the square

You can also find the vertex by rewriting the formula. The part x² − 6x is the start of (x − 3)²:

(x − 3)² = x² − 6x + 9

That has an extra 9 that does not belong. Take it away again:

x² − 6x + 1 = (x − 3)² − 9 + 1 = (x − 3)² − 8

A square is never negative. So (x − 3)² is smallest at x = 3, where it is 0. Then y = −8. The vertex is (3, −8) again.

Vertex form: drag the vertex

The form y = a(x − p)² + q is called vertex form. You read the vertex straight from it: (p, q). Drag the dot below, or tap anywhere on the grid.

Drag the vertex

y = (x − 3)² − 8

• y = a(x − p)² + q
• y = ax², not shifted
• Vertex

a = 1

Vertex

(3, −8)

Turning value

minimum −8

Range

y ≥ −8

Expanded

y = x² − 6x + 1

Show the numbers

x	y
0	1
1	−4
2	−7
3	−8
4	−7
5	−4
6	1

Watch the sign inside the brackets. In (x − 3)², p = 3. In (x + 2)², p = −2, because x + 2 = x − (−2).

The vertex also gives you the minimum or maximum and the range:

• A parabola that opens upward (a positive) has a minimum q. Every y-value is q or more, so the range is y ≥ q.
• A parabola that opens downward (a negative) has a maximum q. The range is y ≤ q.

For y = (x − 3)² − 8 the minimum is −8 and the range is y ≥ −8.

The roots

The roots are the x-values where y = 0. That is where the graph crosses the x-axis. So you solve an equation. Take x² − 4x − 21 = 0.

Way 1: factorising

Find two numbers that multiply to −21 and add up to −4. Those are 3 and −7, because 3 × −7 = −21 and 3 + −7 = −4.

x² − 4x − 21 = 0
(x + 3)(x − 7) = 0
x = −3 or x = 7

A product is 0 when one of its factors is 0. That is why you may set each bracket to 0 on its own. Check: (−3)² − 4 × (−3) − 21 = 9 + 12 − 21 = 0.

The graph shows the same thing. It crosses the x-axis at −3 and at 7. The vertex sits exactly halfway, at x = 2.

Where the graph crosses the x-axis

• y = x² − 4x − 21
• Axis of symmetry x = 2
• Roots
• Vertex

Show the numbers

x	y
−4	11
−3	0
−2	−9
−1	−16
0	−21
1	−24
2	−25
3	−24
4	−21
5	−16
6	−9
7	0
8	11

Way 2: the quadratic formula

Factorising does not always work. The quadratic formula always does. First work out the discriminant:

D = b² − 4ac

Then the roots are:

x = −b + √D2a or x = −b − √D2a

For x² − 4x − 21 = 0, a = 1, b = −4 and c = −21:

D = (−4)² − 4 × 1 × (−21) = 16 + 84 = 100
x = (4 + 10) ÷ 2 = 7 or x = (4 − 10) ÷ 2 = −3

The same answers. Now an equation that does not factorise nicely: x² − 6x + 1 = 0.

D = (−6)² − 4 × 1 × 1 = 36 − 4 = 32
x = (6 + √32) ÷ 2 ≈ 5.83 or x = (6 − √32) ÷ 2 ≈ 0.17

If you want it exact: √32 = √(16 × 2) = 4√2. Then x = 3 + 2√2 or x = 3 − 2√2.

What the discriminant tells you

You do not need to work out the roots to know how many there are. Just look at D:

• D > 0: two roots. The parabola crosses the x-axis twice.
• D = 0: one root. The vertex sits exactly on the x-axis.
• D < 0: no roots. You cannot take the square root of a negative number.

Below you see y = x² − 6x + c three times. Only c changes. The bigger c, the higher the parabola and the smaller D.

Two roots, one or none

y = x² − 6x + c

• Graph
• Axis of symmetry
• Vertex
• Roots

c = 5

D = 16
2 roots

c = 9

D = 0
1 root

c = 13

D = −16
no roots

Show the numbers

c	D	Number of roots
5	16	2
9	0	1
13	−16	0

• c = 5: D = 36 − 20 = 16. Factorising gives x² − 6x + 5 = (x − 1)(x − 5), so x = 1 or x = 5.
• c = 9: D = 36 − 36 = 0. Here x² − 6x + 9 = (x − 3)², and the only root is x = 3.
• c = 13: D = 36 − 52 = −16. The vertex is (3, 4), because y = 9 − 18 + 13 = 4. A parabola that opens upward with its lowest point at height 4 never reaches the x-axis.

Try it in the first chart: set a = 1 and b = −6, then slide c from 5 to 9 and on to 10. The two roots move towards each other, become one, and vanish.

Back to the ball

The ball's height was y = −0.05x² + x. Here a = −0.05, b = 1 and c = 0. Because a is negative, the parabola opens downward.

The vertex:

x = −1 ÷ (2 × −0.05) = −1 ÷ −0.1 = 10
y = −0.05 × 10² + 10 = −5 + 10 = 5

The ball goes 5 metres high, when it is 10 metres away.

Where the ball lands, its height is 0. Take x outside the brackets:

−0.05x² + x = 0
x(−0.05x + 1) = 0
x = 0 or x = 20

At x = 0 the ball is kicked. At x = 20 it hits the ground again. The vertex is at 10, exactly halfway. That is the parabola's symmetry.

Slide the ball along its path. The second ball sits at 20 − x and is always at the same height.

The ball's path

y = −0.05x² + x

• Path
• Ball
• Second ball at 20 − x
• Vertex

Distance: x = 4 m

Height

y = 3.2 m

The ball

goes up

Same height at

x = 16 m

Show the numbers

x (m)	y (m)
0	0
2	1.8
4	3.2
6	4.2
8	4.8
10	5
12	4.8
14	4.2
16	3.2
18	1.8
20	0

Try it yourself