Non-linear graphs - Year 10 Mathematics

What you will learn

sketch parabolas in the form $y = ax^2 + bx + c$ and vertex form $y = a(x - h)^2 + k$ ,
identify the centre and radius of a circle from $(x - h)^2 + (y - k)^2 = r^2$ ,
recognise and sketch exponential curves $y = a^x$ (growth vs decay),
recognise the hyperbola $y = \tfrac{k}{x}$ and its key features.

Worked example 0 Real-world example: height of a kicked football

A football is kicked and its height in metres is modelled by $h = -5t^2 + 15t$ , where $t$ is time in seconds. Find the maximum height.

This is a parabola opening downward ( $a = -5 < 0$ ).
Vertex $t$ -coordinate: $t = -\dfrac{b}{2a} = -\dfrac{15}{2(-5)} = 1.5$ seconds.
Maximum height: $h = -5(1.5)^2 + 15(1.5) = -11.25 + 22.5 = 11.25$ metres.

Key idea: the vertex of a downward parabola gives the maximum value.

1. Parabolas

The graph of any quadratic function is a parabola.

Formula reference

General form: $y = ax^2 + bx + c$ . Vertex at $x = -\dfrac{b}{2a}$ .

Vertex (turning-point) form: $y = a(x - h)^2 + k$ . Vertex at $(h, k)$ .

If $a > 0$ the parabola opens upward (minimum at vertex).
If $a < 0$ the parabola opens downward (maximum at vertex).

Worked example 1 Sketching from vertex form

Sketch $y = 2(x - 3)^2 - 8$ .

Vertex: $(3, -8)$ .
Opens upward ( $a = 2 > 0$ ).
$y$ -intercept: set $x = 0$ : $y = 2(9) - 8 = 10$ , point $(0, 10)$ .
$x$ -intercepts: $0 = 2(x-3)^2 - 8$ , $(x-3)^2 = 4$ , $x - 3 = \pm 2$ , so $x = 1$ or $x = 5$ .
Plot vertex, intercepts, and draw a smooth U-shape.

Worked example 2 Converting general to vertex form

Write $y = x^2 - 6x + 5$ in vertex form.

Complete the square: $y = (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4$ .
Vertex: $(3, -4)$ . The parabola opens upward.

2. Circles

Formula reference

(x - h)^2 + (y - k)^2 = r^2

Centre $(h, k)$ , radius $r$ .

Worked example 3 Identifying centre and radius

Find the centre and radius of $(x + 2)^2 + (y - 5)^2 = 49$ .

Rewrite: $(x - (-2))^2 + (y - 5)^2 = 7^2$ .
Centre $(-2, 5)$ , radius $7$ .

Worked example 4 Writing the equation of a circle

Write the equation of a circle with centre $(3, -1)$ and radius $4$ .

$(x - 3)^2 + (y + 1)^2 = 16$ .

3. Exponential functions

Formula reference

y = a^x

If $a > 1$ : exponential growth (curve rises steeply to the right).
If $0 < a < 1$ : exponential decay (curve falls toward zero).
The $y$ -intercept is always $(0, 1)$ (since $a^0 = 1$ ).
The $x$ -axis ( $y = 0$ ) is a horizontal asymptote.

Worked example 5 Sketching an exponential growth curve

Sketch $y = 2^x$ .

Key points: $(-2, \tfrac{1}{4})$ , $(-1, \tfrac{1}{2})$ , $(0, 1)$ , $(1, 2)$ , $(2, 4)$ , $(3, 8)$ .
Curve passes through $(0, 1)$ , rises steeply to the right.
As $x \to -\infty$ , $y \to 0$ (approaches the $x$ -axis but never touches it).

4. Hyperbolas (reciprocal functions)

Formula reference

y = \frac{k}{x}

Two branches: one in the first and third quadrants if $k > 0$ ; second and fourth quadrants if $k < 0$ .
Asymptotes: the $x$ -axis ( $y = 0$ ) and the $y$ -axis ( $x = 0$ ).
The graph never crosses either axis.

Worked example 6 Sketching a hyperbola

Sketch $y = \dfrac{6}{x}$ .

Key points: $(1, 6)$ , $(2, 3)$ , $(3, 2)$ , $(6, 1)$ , $(-1, -6)$ , $(-2, -3)$ .
$k = 6 > 0$ : branches in quadrants I and III.
As $x \to \infty$ , $y \to 0$ . As $x \to 0^+$ , $y \to \infty$ .

Three non-linear graphs on separate mini-axes: a parabola y = x squared minus 4, a circle centred at (0, 0) with radius 3, and an exponential y = 2 to the x.

Practice

Fluency

Tier 1: basic skills

State the vertex and direction (up/down) of $y = (x + 1)^2 - 9$ .
Find the $x$ -intercepts of $y = x^2 - 4x - 5$ by factorising.
State the centre and radius of $(x - 3)^2 + (y + 2)^2 = 25$ .
Write the equation of a circle with centre $(0, 4)$ and radius $6$ .
For $y = 3^x$ , find $y$ when $x = 0$ , $x = 2$ , and $x = -1$ .
For $y = \dfrac{4}{x}$ , find $y$ when $x = 1$ , $x = 4$ , and $x = -2$ .
Identify whether each function is a parabola, circle, exponential, or hyperbola: (a) $y = 5^x$ , (b) $x^2 + y^2 = 16$ , (c) $y = \tfrac{3}{x}$ , (d) $y = -x^2 + 2$ .
State the asymptote(s) of $y = \dfrac{-2}{x}$ .

Reasoning

Tier 2: mixed practice

Write $y = x^2 + 8x + 12$ in vertex form by completing the square, then state the vertex.
A ball is thrown upward with height $h = -4.9t^2 + 19.6t + 1.5$ . Find the maximum height and the time it is reached.
Determine whether the point $(3, 4)$ lies inside, on, or outside the circle $(x - 1)^2 + (y - 2)^2 = 9$ .
Sketch $y = 2^x$ and $y = \left(\tfrac{1}{2}\right)^x$ on the same axes. Describe the relationship between the two curves.
A hyperbola passes through $(2, 5)$ and has the form $y = \tfrac{k}{x}$ . Find $k$ and state the equations of the asymptotes.
The parabola $y = a(x - 1)^2 + 3$ passes through $(3, 11)$ . Find $a$ .

Reasoning

Tier 3: explain and apply

A tunnel has a parabolic cross-section. At ground level it is $8$ m wide and the maximum height is $5$ m. Taking the origin at the centre of the base, find the equation of the parabola and determine whether a truck $3$ m wide and $4$ m tall can pass through.
Show algebraically that the circle $x^2 + y^2 = 25$ and the line $y = x + 1$ intersect at two points. Find the coordinates.
Compare the graphs of $y = 2^x$ and $y = 3 \times 2^x$ . Explain how the multiplier affects the curve.
Explain why $y = \dfrac{k}{x}$ can never equal zero, no matter how large $x$ becomes.

Challenge

Reasoning

Harder reasoning

A quadratic $y = ax^2 + bx + c$ has vertex $(2, -3)$ and passes through $(0, 5)$ . Find $a$ , $b$ , and $c$ .
Find the two points where the parabola $y = x^2$ and the circle $x^2 + y^2 = 2$ intersect (for $y \geq 0$ ).
The curve $y = 2^x$ is reflected in the $y$ -axis and then shifted up by $3$ units. Write the equation of the resulting curve and state its asymptote.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Year 10 core - answers

Fluency

Tier 1: basic skills

Vertex $(-1, -9)$ . Opens upward ( $a = 1 > 0$ ).
$x^2 - 4x - 5 = (x - 5)(x + 1) = 0$ . $x$ -intercepts: $x = 5$ and $x = -1$ .
Centre $(3, -2)$ , radius $5$ .
$x^2 + (y - 4)^2 = 36$ .
$x = 0$ : $y = 1$ . $x = 2$ : $y = 9$ . $x = -1$ : $y = \tfrac{1}{3}$ .
$x = 1$ : $y = 4$ . $x = 4$ : $y = 1$ . $x = -2$ : $y = -2$ .
(a) Exponential. (b) Circle. (c) Hyperbola. (d) Parabola.
Horizontal asymptote: $y = 0$ . Vertical asymptote: $x = 0$ .

Reasoning

Tier 2: mixed practice

$y = (x^2 + 8x + 16) - 16 + 12 = (x + 4)^2 - 4$ . Vertex $(-4, -4)$ .
$t = -\dfrac{19.6}{2(-4.9)} = 2$ seconds. $h = -4.9(4) + 19.6(2) + 1.5 = -19.6 + 39.2 + 1.5 = 21.1$ metres.
Substitute: $(3-1)^2 + (4-2)^2 = 4 + 4 = 8$ . Since $8 < 9 = r^2$ , the point is inside the circle.
$y = \left(\tfrac{1}{2}\right)^x = 2^{-x}$ , so the second curve is a reflection of $y = 2^x$ in the $y$ -axis. Both pass through $(0, 1)$ . One grows right, the other decays right.
$5 = \tfrac{k}{2}$ , so $k = 10$ . Asymptotes: $x = 0$ and $y = 0$ .
$11 = a(3-1)^2 + 3 = 4a + 3$ . $4a = 8$ . $a = 2$ .

Reasoning

Tier 3: explain and apply

The base goes from $(-4, 0)$ to $(4, 0)$ . Maximum height at $(0, 5)$ . Equation: $y = a x^2 + 5$ . At $x = 4$ : $0 = 16a + 5$ , so $a = -\tfrac{5}{16}$ . Equation: $y = -\tfrac{5}{16}x^2 + 5$ . The truck is $3$ m wide, so its edges are at $x = \pm 1.5$ . Height at $x = 1.5$ : $y = -\tfrac{5}{16}(2.25) + 5 = -0.703 + 5 = 4.297$ m. Since $4.297 > 4$ , yes, the truck can pass through.
Substitute $y = x + 1$ into $x^2 + y^2 = 25$ : $x^2 + (x+1)^2 = 25$ . $2x^2 + 2x + 1 = 25$ . $2x^2 + 2x - 24 = 0$ . $x^2 + x - 12 = 0$ . $(x+4)(x-3) = 0$ . $x = -4$ or $x = 3$ . Points: $(-4, -3)$ and $(3, 4)$ .
$y = 3 \times 2^x$ has the same shape as $y = 2^x$ but every $y$ -value is multiplied by $3$ . It is a vertical stretch by factor $3$ . The $y$ -intercept moves from $(0, 1)$ to $(0, 3)$ . The asymptote remains $y = 0$ .
For $y = \tfrac{k}{x}$ to equal zero, we would need $\tfrac{k}{x} = 0$ , which means $k = 0 \times x = 0$ . But $k \neq 0$ (otherwise it is not a hyperbola), so no value of $x$ can make $y = 0$ . As $x$ grows larger, $y$ gets closer and closer to $0$ but never reaches it.

Reasoning

Challenge

Vertex form: $y = a(x - 2)^2 - 3$ . Through $(0, 5)$ : $5 = a(4) - 3$ , $a = 2$ . Expand: $y = 2(x^2 - 4x + 4) - 3 = 2x^2 - 8x + 5$ . So $a = 2$ , $b = -8$ , $c = 5$ .
From $y = x^2$ , substitute into $x^2 + y^2 = 2$ : $y + y^2 = 2$ (since $x^2 = y$ ). $y^2 + y - 2 = 0$ . $(y+2)(y-1) = 0$ . $y = 1$ (take $y \geq 0$ ). Then $x^2 = 1$ , so $x = \pm 1$ . Points: $(1, 1)$ and $(-1, 1)$ .
Reflect in $y$ -axis: $y = 2^{-x}$ . Shift up $3$ : $y = 2^{-x} + 3$ . Asymptote: $y = 3$ .

Prefer paper? Print the answer key as a separate booklet: open print view ->