Year 10 Mathematics | Victorian Curriculum 2.0
Non-linear graphs
Topic 06 | Number & Algebra | Practice

What you will learn

  • sketch parabolas in the form y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c and vertex form y=a(x−h)2+ky = a(x - h)^2 + ky=a(x−h)2+k,
  • identify the centre and radius of a circle from (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2(x−h)2+(y−k)2=r2,
  • recognise and sketch exponential curves y=axy = a^xy=ax (growth vs decay),
  • recognise the hyperbola y=kxy = \tfrac{k}{x}y=xk​ and its key features.
Why study non-linear graphs?

The real world is rarely perfectly linear. Projectiles follow parabolas, populations grow exponentially, radio signals weaken as 1/r21/r^21/r2, and circles appear everywhere from wheels to orbits. Being able to read and sketch these curves lets you model a much wider range of phenomena.

Where you'll see this
  • Sport: the arc of a ball or javelin is a parabola.
  • Medicine: bacterial growth follows an exponential curve.
  • Engineering: circular cross-sections in pipes, tunnels, and gears.
  • Physics: the brightness of light decreases as a reciprocal of distance squared.
Worked example 0 Real-world example: height of a kicked football

A football is kicked and its height in metres is modelled by h=−5t2+15th = -5t^2 + 15th=−5t2+15t, where ttt is time in seconds. Find the maximum height.

  1. This is a parabola opening downward (a=−5<0a = -5 < 0a=−5<0).
  2. Vertex ttt-coordinate: t=−b2a=−152(−5)=1.5t = -\dfrac{b}{2a} = -\dfrac{15}{2(-5)} = 1.5t=−2ab​=−2(−5)15​=1.5 seconds.
  3. Maximum height: h=−5(1.5)2+15(1.5)=−11.25+22.5=11.25h = -5(1.5)^2 + 15(1.5) = -11.25 + 22.5 = 11.25h=−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=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c. Vertex at x=−b2ax = -\dfrac{b}{2a}x=−2ab​.

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

  • If a>0a > 0a>0 the parabola opens upward (minimum at vertex).
  • If a<0a < 0a<0 the parabola opens downward (maximum at vertex).
Worked example 1 Sketching from vertex form

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

  1. Vertex: (3,−8)(3, -8)(3,−8).
  2. Opens upward (a=2>0a = 2 > 0a=2>0).
  3. yyy-intercept: set x=0x = 0x=0: y=2(9)−8=10y = 2(9) - 8 = 10y=2(9)−8=10, point (0,10)(0, 10)(0,10).
  4. xxx-intercepts: 0=2(x−3)2−80 = 2(x-3)^2 - 80=2(x−3)2−8, (x−3)2=4(x-3)^2 = 4(x−3)2=4, x−3=±2x - 3 = \pm 2x−3=±2, so x=1x = 1x=1 or x=5x = 5x=5.
  5. Plot vertex, intercepts, and draw a smooth U-shape.
Worked example 2 Converting general to vertex form

Write y=x2−6x+5y = x^2 - 6x + 5y=x2−6x+5 in vertex form.

  1. Complete the square: y=(x2−6x+9)−9+5=(x−3)2−4y = (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4y=(x2−6x+9)−9+5=(x−3)2−4.
  2. Vertex: (3,−4)(3, -4)(3,−4). The parabola opens upward.

2. Circles

Formula reference

(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2(x−h)2+(y−k)2=r2

Centre (h,k)(h, k)(h,k), radius rrr.

Worked example 3 Identifying centre and radius

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

  1. Rewrite: (x−(−2))2+(y−5)2=72(x - (-2))^2 + (y - 5)^2 = 7^2(x−(−2))2+(y−5)2=72.
  2. Centre (−2,5)(-2, 5)(−2,5), radius 777.
Worked example 4 Writing the equation of a circle

Write the equation of a circle with centre (3,−1)(3, -1)(3,−1) and radius 444.

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

3. Exponential functions

Formula reference

y=axy = a^xy=ax
  • If a>1a > 1a>1: exponential growth (curve rises steeply to the right).
  • If 0<a<10 < a < 10<a<1: exponential decay (curve falls toward zero).
  • The yyy-intercept is always (0,1)(0, 1)(0,1) (since a0=1a^0 = 1a0=1).
  • The xxx-axis (y=0y = 0y=0) is a horizontal asymptote.
Worked example 5 Sketching an exponential growth curve

Sketch y=2xy = 2^xy=2x.

  1. Key points: (−2,14)(-2, \tfrac{1}{4})(−2,41​), (−1,12)(-1, \tfrac{1}{2})(−1,21​), (0,1)(0, 1)(0,1), (1,2)(1, 2)(1,2), (2,4)(2, 4)(2,4), (3,8)(3, 8)(3,8).
  2. Curve passes through (0,1)(0, 1)(0,1), rises steeply to the right.
  3. As x→−∞x \to -\inftyx→−∞, y→0y \to 0y→0 (approaches the xxx-axis but never touches it).

4. Hyperbolas (reciprocal functions)

Formula reference

y=kxy = \frac{k}{x}y=xk​
  • Two branches: one in the first and third quadrants if k>0k > 0k>0; second and fourth quadrants if k<0k < 0k<0.
  • Asymptotes: the xxx-axis (y=0y = 0y=0) and the yyy-axis (x=0x = 0x=0).
  • The graph never crosses either axis.
Worked example 6 Sketching a hyperbola

Sketch y=6xy = \dfrac{6}{x}y=x6​.

  1. Key points: (1,6)(1, 6)(1,6), (2,3)(2, 3)(2,3), (3,2)(3, 2)(3,2), (6,1)(6, 1)(6,1), (−1,−6)(-1, -6)(−1,−6), (−2,−3)(-2, -3)(−2,−3).
  2. k=6>0k = 6 > 0k=6>0: branches in quadrants I and III.
  3. As x→∞x \to \inftyx→∞, y→0y \to 0y→0. As x→0+x \to 0^+x→0+, y→∞y \to \inftyy→∞.
Parabolaxy(0,-4)Circlexyr=3Exponentialxy(0,1)
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

    1. State the vertex and direction (up/down) of y=(x+1)2−9y = (x + 1)^2 - 9y=(x+1)2−9.
    2. Find the xxx-intercepts of y=x2−4x−5y = x^2 - 4x - 5y=x2−4x−5 by factorising.
    3. State the centre and radius of (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25(x−3)2+(y+2)2=25.
    4. Write the equation of a circle with centre (0,4)(0, 4)(0,4) and radius 666.
    5. For y=3xy = 3^xy=3x, find yyy when x=0x = 0x=0, x=2x = 2x=2, and x=−1x = -1x=−1.
    6. For y=4xy = \dfrac{4}{x}y=x4​, find yyy when x=1x = 1x=1, x=4x = 4x=4, and x=−2x = -2x=−2.
    7. Identify whether each function is a parabola, circle, exponential, or hyperbola: (a) y=5xy = 5^xy=5x, (b) x2+y2=16x^2 + y^2 = 16x2+y2=16, (c) y=3xy = \tfrac{3}{x}y=x3​, (d) y=−x2+2y = -x^2 + 2y=−x2+2.
    8. State the asymptote(s) of y=−2xy = \dfrac{-2}{x}y=x−2​.
Reasoning

Tier 2: mixed practice

    1. Write y=x2+8x+12y = x^2 + 8x + 12y=x2+8x+12 in vertex form by completing the square, then state the vertex.
    2. A ball is thrown upward with height h=−4.9t2+19.6t+1.5h = -4.9t^2 + 19.6t + 1.5h=−4.9t2+19.6t+1.5. Find the maximum height and the time it is reached.
    3. Determine whether the point (3,4)(3, 4)(3,4) lies inside, on, or outside the circle (x−1)2+(y−2)2=9(x - 1)^2 + (y - 2)^2 = 9(x−1)2+(y−2)2=9.
    4. Sketch y=2xy = 2^xy=2x and y=(12)xy = \left(\tfrac{1}{2}\right)^xy=(21​)x on the same axes. Describe the relationship between the two curves.
    5. A hyperbola passes through (2,5)(2, 5)(2,5) and has the form y=kxy = \tfrac{k}{x}y=xk​. Find kkk and state the equations of the asymptotes.
    6. The parabola y=a(x−1)2+3y = a(x - 1)^2 + 3y=a(x−1)2+3 passes through (3,11)(3, 11)(3,11). Find aaa.
Reasoning

Tier 3: explain and apply

    1. A tunnel has a parabolic cross-section. At ground level it is 888 m wide and the maximum height is 555 m. Taking the origin at the centre of the base, find the equation of the parabola and determine whether a truck 333 m wide and 444 m tall can pass through.
    2. Show algebraically that the circle x2+y2=25x^2 + y^2 = 25x2+y2=25 and the line y=x+1y = x + 1y=x+1 intersect at two points. Find the coordinates.
    3. Compare the graphs of y=2xy = 2^xy=2x and y=3×2xy = 3 \times 2^xy=3×2x. Explain how the multiplier affects the curve.
    4. Explain why y=kxy = \dfrac{k}{x}y=xk​ can never equal zero, no matter how large xxx becomes.

Challenge

Reasoning

Harder reasoning

    1. A quadratic y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c has vertex (2,−3)(2, -3)(2,−3) and passes through (0,5)(0, 5)(0,5). Find aaa, bbb, and ccc.
    2. Find the two points where the parabola y=x2y = x^2y=x2 and the circle x2+y2=2x^2 + y^2 = 2x2+y2=2 intersect (for y≥0y \geq 0y≥0).
    3. The curve y=2xy = 2^xy=2x is reflected in the yyy-axis and then shifted up by 333 units. Write the equation of the resulting curve and state its asymptote.
Year 10 Mathematics study companion | Practice