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Functions, relations and graphs: VCE Math Methods Unit 1 Year 11

Syllabus dot point

“Linear, quadratic, cubic and quartic polynomial functions, basic exponential functions y=axy = a^x, logarithmic functions y=log⁡a(x)y = \log_a(x), and the standard transformations (dilation, reflection, translation)”

8 min read

Quick answer

Unit 1 introduces the major function families (linear, quadratic, polynomial up to quartic, exponential, logarithmic) and the four standard transformations (translation in xx and yy, dilation in xx and yy); sketching requires all key features (intercepts, turning points, asymptotes, end behaviour) labelled, and transformations applied in the correct order (inside-the-bracket first, outside second).

Jump to a section
  1. What this dot point is asking
  2. Function families
  3. Key graphical features
  4. The four transformations
  5. Domain, range and inverse
  6. Examples in context
  7. Try this
  8. Exam-style questions

What this dot point is asking

VCAA wants you to recognise the standard function families introduced in Unit 1, identify their key graphical features (intercepts, turning points, asymptotes), and apply the four standard transformations (dilation, reflection, translation in xx and yy).

Function families

Linear y=mx+cy = mx + c. Gradient mm, yy-intercept cc. Sketched as a straight line.

Quadratic y=ax2+bx+cy = a x^2 + bx + c or vertex form y=a(x−h)2+ky = a(x-h)^2 + k. Parabola opening up if a>0a > 0, down if a<0a < 0. Turning point at (h,k)(h, k) in vertex form.

Cubic y=ax3+bx2+cx+dy = a x^3 + b x^2 + c x + d. Either monotonic or with two turning points. Inflection at the average of the turning points.

Quartic y=ax4+…y = a x^4 + \ldots. Either two or four sign changes; up to three turning points.

Exponential y=axy = a^x for a>0,a≠1a > 0, a \neq 1. Always positive. Horizontal asymptote y=0y = 0. Passes through (0,1)(0, 1). Increasing if a>1a > 1, decreasing if 0<a<10 < a < 1.

Logarithmic y=log⁡a(x)y = \log_a(x) for a>0,a≠1a > 0, a \neq 1. Defined only for x>0x > 0. Vertical asymptote x=0x = 0. Passes through (1,0)(1, 0). Inverse of y=axy = a^x.

Key graphical features

For each function:

  • Domain. Set of allowed xx values.
  • Range. Set of resulting yy values.
  • Axis intercepts. Where the graph crosses the xx- and yy-axes.
  • Turning points / stationary points. Local maxima and minima.
  • Asymptotes. Lines the graph approaches but never meets.
  • End behaviour. What happens as x→±∞x \to \pm \infty.

Sketching requires all relevant features labelled.

The four transformations

Given y=f(x)y = f(x), the transformations:

Translation in yy
y=f(x)+ky = f(x) + k shifts up by kk (down if k<0k < 0).
Translation in xx
y=f(x−h)y = f(x - h) shifts right by hh (left if h<0h < 0). Note the sign convention: (x−h)(x - h) means shift right.
Dilation in yy
y=af(x)y = a f(x) stretches vertically by factor aa (compresses if 0<a<10 < a < 1, reflects if a<0a < 0).
Dilation in xx
y=f(bx)y = f(b x) stretches horizontally by factor 1/b1/b (compresses if b>1b > 1).
Combined transformations
y=af(b(x−h))+ky = a f(b(x - h)) + k combines all four with vertex at (h,k)(h, k).
Reflections
Special case of dilations:
  • y=−f(x)y = -f(x): reflection in xx-axis.
  • y=f(−x)y = f(-x): reflection in yy-axis.
Worked example

Step 1: Sketch the linear function

The function y=2x−3y = 2x - 3 has the form y=mx+cy = mx + c, where the gradient m=2m = 2 tells us the line rises two units for every one unit across, and the yy-intercept c=−3c = -3 gives the point where it crosses the vertical axis. To find the xx-intercept, set y=0y = 0 and solve:

0=2x−3  ⟹  x=1.50 = 2x - 3 \implies x = 1.5

Sketch the straight line through the two known points (0,−3)(0, -3) and (1.5,0)(1.5, 0).

Step 2: Identify transformations on the quadratic

Start from the base parabola y=x2y = x^2, which has its vertex at the origin. The transformed form y=2(x−1)2−5y = 2(x - 1)^2 - 5 encodes three separate changes: the factor 22 outside the bracket dilates the graph vertically (making it steeper), the −1-1 inside the bracket shifts the vertex one unit to the right, and the −5-5 outside shifts it five units down. The vertex is therefore at (1,−5)(1, -5) and the parabola opens upward because the coefficient 22 is positive.

Step 3: Identify transformations on the exponential

Begin with the base curve y=3xy = 3^x, which has a horizontal asymptote at y=0y = 0 and passes through (0,1)(0, 1). The function y=2⋅3x+1y = 2 \cdot 3^x + 1 applies a vertical dilation by factor 22 (stretching away from the xx-axis), then a vertical translation of +1+1 (shifting the entire curve upward). The new horizontal asymptote moves to y=1y = 1, and the yy-intercept becomes 2⋅30+1=2⋅1+1=32 \cdot 3^0 + 1 = 2 \cdot 1 + 1 = 3.

Final answer: The three key features for each graph are: (1) line through (0,−3)(0, -3) and (1.5,0)(1.5, 0); (2) parabola with vertex (1,−5)(1, -5) opening upward; (3) exponential with asymptote y=1y = 1 and yy-intercept 33.

Domain, range and inverse

For inverse functions (covered in Unit 4), the domain and range swap. For Unit 1, observe:

  • f(x)=axf(x) = a^x has domain R\mathbb{R}, range (0,∞)(0, \infty).
  • f−1(x)=log⁡a(x)f^{-1}(x) = \log_a(x) has domain (0,∞)(0, \infty), range R\mathbb{R}.

The domains and ranges are reflections in the line y=xy = x.

Common errors
Translation sign error
y=f(x−3)y = f(x - 3) shifts right by 3 (positive direction), not left.
Wrong order of transformations
Apply inside-the-bracket transformations first (operations on xx), then outside (operations on yy).
Forgetting the asymptote on exponential / log graphs
Exponentials have horizontal asymptotes; logs have vertical asymptotes. Mark them.
Confusing domain with range
Domain is the set of valid inputs; range is the set of outputs. For logs, the domain is restricted to positive xx.

Examples in context

Example 1. Transforming a quadratic. Starting from y=x2y = x^2, the graph of y=2(x−3)2−5y = 2(x - 3)^2 - 5 is dilated vertically by factor 22, translated 33 right and 55 down. Its vertex is at (3,−5)(3, -5), and the yy-intercept is y=2(0−3)2−5=18−5=13y = 2(0 - 3)^2 - 5 = 18 - 5 = 13. The parabola opens upward.

Example 2. Exponential vs logarithmic features. The graph of y=2xy = 2^x has horizontal asymptote y=0y = 0, yy-intercept 11, and range y>0y > 0. Its inverse y=log⁡2xy = \log_2 x has vertical asymptote x=0x = 0, xx-intercept 11, and domain x>0x > 0. They are reflections of each other in the line y=xy = x.

Try this

Q1. Describe the transformations taking y=x2y = x^2 to y=(x+1)2+4y = (x + 1)^2 + 4, and state the vertex. [2 marks]

  • Cue. Translate 11 left and 44 up; vertex (−1,4)(-1, 4).

Q2. State the asymptote, yy-intercept and range of y=3x−2y = 3^x - 2. [3 marks]

  • Cue. Asymptote y=−2y = -2; yy-intercept 30−2=−13^0 - 2 = -1; range y>−2y > -2.

Q3. The graph of y=a(x−h)2+ky = a(x - h)^2 + k has vertex (2,−1)(2, -1) and passes through (0,3)(0, 3). Find aa. [2 marks]

  • Cue. 3=a(0−2)2−1=4a−13 = a(0 - 2)^2 - 1 = 4a - 1, so a=1a = 1.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Year 11 SAC-style4 marks
Sketch the graph of y=−2(x−1)2+3y = -2(x-1)^2 + 3, labelling the turning point, yy-intercept and any xx-intercepts.
Show worked answer →

Quadratic in vertex form y=a(x−h)2+ky = a(x - h)^2 + k has turning point (h,k)=(1,3)(h, k) = (1, 3) and opens downward because a=−2<0a = -2 < 0.

yy-intercept: substitute x=0x = 0. y=−2(0−1)2+3=−2+3=1y = -2(0-1)^2 + 3 = -2 + 3 = 1. Intercept: (0,1)(0, 1).

xx-intercepts: set y=0y = 0. −2(x−1)2+3=0-2(x-1)^2 + 3 = 0, so (x−1)2=3/2(x-1)^2 = 3/2, giving x−1=±3/2x - 1 = \pm \sqrt{3/2}, i.e. x=1±6/2≈1±1.22x = 1 \pm \sqrt{6}/2 \approx 1 \pm 1.22. Intercepts: approximately (−0.22,0)(-0.22, 0) and (2.22,0)(2.22, 0).

Sketch: downward parabola with turning point at (1,3)(1, 3), yy-intercept (0,1)(0, 1), xx-intercepts at the calculated points.

Markers reward correct turning point, intercepts (exact values preferred where possible), and shape.

Practise this

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