Open Math›Year 10›Non-linear relationships›Parabolas

Parabolas

Identify the vertex, axis of symmetry and concavity of parabolas, describe the effect of changing coefficients, and find x-intercepts by solving y = 0.

Worked examples

Identifying vertex, axis of symmetry and concavity

Straightforward

Problem

For the parabola

y = -3x^2 + 4

, state: (a) the vertex, (b) the equation of the axis of symmetry, and (c) whether it opens upward or downward.

Identify the values of

k

and

c

y = kx^2 + c

Comparing

y = -3x^2 + 4

with

y = kx^2 + c

k = -3

and

c = 4

State the vertex using

(0, c)

The vertex of any parabola of the form

y = kx^2 + c

is at

(0, c)

. Here the vertex is

(0, 4)

. Check: when

x = 0

y = -3(0)^2 + 4 = 4

✓

State the axis of symmetry.

The axis of symmetry is the vertical line through the vertex. Since the vertex is at

(0, 4)

, the axis of symmetry is

x = 0

(the y-axis).

Determine whether the parabola opens upward or downward.

The coefficient of

x^2

k = -3

. Since

k < 0

, the parabola opens downward. The vertex

(0, 4)

is the maximum point.

Answer

Vertex

(0, 4)

; axis of symmetry

x = 0

; opens downward.

Describing the effect of changing $k$ and $c$

Moderate

Problem

Compare the parabolas

y = x^2

y = 3x^2

, and

y = x^2 - 5

. (a) Which is narrower:

y = x^2

y = 3x^2

? (b) How does

y = x^2 - 5

differ from

y = x^2

Compare

y = x^2

and

y = 3x^2

by examining the coefficient of

x^2

y = x^2

has coefficient

k = 1

y = 3x^2

has coefficient

k = 3

. At

x = 2

y = 4

for

y = x^2

but

y = 12

for

y = 3x^2

— the second parabola rises more steeply.

State which parabola is narrower.

Since

3 > 1

y = 3x^2

is narrower than

y = x^2

. Both have vertex

(0, 0)

and axis of symmetry

x = 0

, but

y = 3x^2

is steeper.

Compare

y = x^2

and

y = x^2 - 5

by examining the constant term.

y = x^2

has vertex

(0, 0)

y = x^2 - 5

has

c = -5

, so its vertex is at

(0, -5)

. Subtracting 5 shifts every point on the graph 5 units downward.

Summarise the effect.

The shape and width of

y = x^2 - 5

are identical to

y = x^2

. The only difference is that the entire graph has shifted 5 units downward.

Answer

(a)

y = 3x^2

is narrower. (b)

y = x^2 - 5

is the graph of

y = x^2

shifted 5 units downward.

Finding the x-intercepts of a parabola

Challenging

Problem

Find the x-intercepts of

y = 3x^2 - 27

Set

y = 0

to find where the parabola crosses the x-axis.

3x^2 - 27 = 0

Isolate

x^2

3x^2 = 27

x^2 = 9

Solve for

x

by taking the square root of both sides.

x = \pm\sqrt{9} = \pm 3

State the x-intercepts as coordinates and interpret the result.

The x-intercepts are

(3, 0)

and

(-3, 0)

. Since

k = 3 > 0

, the parabola opens upward with vertex at

(0, -27)

below the x-axis, confirming it crosses the x-axis at two points.

Verify by substituting back into the original equation.

When

x = 3

y = 3(3)^2 - 27 = 27 - 27 = 0

✓ When

x = -3

y = 3(-3)^2 - 27 = 27 - 27 = 0

✓

Answer

The x-intercepts are

x = 3

and

x = -3

, at the points

(3, 0)

and

(-3, 0)

Practise

Q1·Straightforward

What is the vertex of the parabola

y = 3x^2 + 5

Explanation

The equation

y = 3x^2 + 5

is in the form

y = kx^2 + c

with

k = 3

and

c = 5

. The vertex is at

(0, c) = (0, 5)

. When

x = 0

y = 3(0)^2 + 5 = 5

✓

Q2·Straightforward

What is the y-coordinate of the vertex of

y = -x^2 - 4

Explanation

y = -x^2 - 4

, the constant term is

c = -4

. The vertex is at

(0, -4)

, so its y-coordinate is

-4

. When

x = 0

y = -(0)^2 - 4 = -4

✓

Q3·Straightforward

Does the parabola

y = -2x^2 + 7

open upward or downward?

UpwardDownwardIt is a horizontal lineIt cannot be determined

Explanation

y = -2x^2 + 7

, the coefficient of

x^2

k = -2

. Since

k < 0

, the parabola opens downward. The vertex

(0, 7)

is the highest point on the curve.

Q4·Straightforward

What is the axis of symmetry of

y = 6x^2 - 3

$x = 0$$x = 6$$x = -3$$y = -3$

Explanation

The parabola

y = 6x^2 - 3

has vertex at

(0, -3)

. The axis of symmetry is the vertical line through the vertex:

x = 0

(the y-axis). The values

x = 6

and

x = -3

come from misreading the coefficients, and

y = -3

is a horizontal line — not an axis of symmetry.

Q5·Moderate

How does the graph of

y = x^2 + 4

differ from the graph of

y = x^2

The graph shifts 4 units upwardThe graph shifts 4 units to the rightThe graph becomes 4 times narrowerThe graph reflects about the x-axis

Explanation

Changing

y = x^2

y = x^2 + 4

adds

c = 4

to the equation, shifting the vertex from

(0, 0)

(0, 4)

— that is, 4 units upward. The shape and width of the parabola are unchanged. Changing the coefficient of

x^2

would affect the width; changing the sign of

x^2

would cause a reflection.

Q6·Moderate

Which parabola is narrower:

y = x^2

y = 5x^2

$y = x^2$$y = 5x^2$They are equally wideBoth are the same width because they share the same vertex

Explanation

y = x^2

, the coefficient is

k = 1

; in

y = 5x^2

, it is

k = 5

. A larger coefficient causes the graph to rise more steeply for the same

x

-value, making it narrower. At

x = 1

y = 1

for

y = x^2

but

y = 5

for

y = 5x^2

, confirming

y = 5x^2

is steeper and narrower.

Q7·Moderate

What is the y-intercept of the parabola

y = -3x^2 + 9

Explanation

Substituting

x = 0

y = -3(0)^2 + 9 = 0 + 9 = 9

The y-intercept is

9

. For any parabola of the form

y = kx^2 + c

, the y-intercept is always

c

(the constant term).

Q8·Moderate

The vertex of the parabola

y = 2x^2 + k

is at

(0, -3)

. What is the value of

k

Explanation

The parabola

y = 2x^2 + k

is in the form

y = ax^2 + c

, so its vertex is at

(0, k)

. Since the vertex is at

(0, -3)

, we need

k = -3

. Verify:

y = 2x^2 - 3

. When

x = 0

y = -3

✓

Q9·Challenging

Find the x-intercepts of

y = x^2 - 36

by setting

y = 0

. What is the positive x-intercept?

Explanation

Setting

y = 0

x^2 - 36 = 0

x^2 = 36

x = \pm\sqrt{36} = \pm 6

The x-intercepts are

(6, 0)

and

(-6, 0)

. The positive x-intercept is

6

Q10·Challenging

Find the x-intercepts of

y = 2x^2 - 8

by setting

y = 0

$x = 2$ and $x = -2$$x = 4$ and $x = -4$$x = 2$ only$x = \sqrt{8}$ and $x = -\sqrt{8}$

Explanation

Setting

y = 0

2x^2 - 8 = 0

2x^2 = 8

x^2 = 4

x = \pm 2

The x-intercepts are

(2, 0)

and

(-2, 0)

. A common error is to treat

2x^2 = 8

2x = 8

, giving

x = 4

only; another is to forget the negative root; a third is to skip dividing by

2

, giving

x^2 = 8

and

x = \pm\sqrt{8}

Q11·Challenging

Which of the following parabolas is the narrowest?

$y = x^2$$y = 2x^2$$y = \dfrac{1}{2}x^2$$y = 4x^2$

Explanation

The coefficients of

x^2

are:

y = x^2

(

k = 1

y = 2x^2

(

k = 2

y = \frac{1}{2}x^2

(

k = 0.5

y = 4x^2

(

k = 4

). Since

4 > 2 > 1 > 0.5

, the parabola

y = 4x^2

is the narrowest. At

x = 1

: its y-value is

4

, compared to

2

1

and

0.5

for the others, confirming it rises the most steeply.

Q12·Challenging

Find the x-intercepts of

y = -x^2 + 49

by setting

y = 0

. What is the positive x-intercept?

Explanation

Setting

y = 0

-x^2 + 49 = 0

-x^2 = -49

x^2 = 49

x = \pm 7

The x-intercepts are

(7, 0)

and

(-7, 0)

. The positive x-intercept is

7

. Since

k = -1 < 0

this parabola opens downward, with vertex

(0, 49)

above the x-axis — so it does cross the x-axis at two points.