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VICMath MethodsSyllabus dot point

What functions and relations are introduced in VCE Math Methods Unit 1, and how are they graphed and transformed?

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

A focused answer to the VCE Math Methods Unit 1 key-knowledge point on functions and graphs. Linear, quadratic, polynomial, exponential and logarithmic functions, their key features (axes intercepts, turning points, asymptotes), and the four standard transformations that prepare for Unit 3 graphical work.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  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

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(xh)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,a1a > 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=loga(x)y = \log_a(x) for a>0,a1a > 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(xh)y = f(x - h) shifts right by hh (left if h<0h < 0). Note the sign convention: (xh)(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(xh))+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.

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).
  • f1(x)=loga(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.

Examples in context

Example 1. Transforming a quadratic. Starting from y=x2y = x^2, the graph of y=2(x3)25y = 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(03)25=185=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=log2xy = \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=3x2y = 3^x - 2. [3 marks]

  • Cue. Asymptote y=2y = -2; yy-intercept 302=13^0 - 2 = -1; range y>2y > -2.

Q3. The graph of y=a(xh)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(02)21=4a13 = a(0 - 2)^2 - 1 = 4a - 1, so a=1a = 1.

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Year 11 SAC4 marksSketch the graph of y=2(x1)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(xh)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(01)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(x1)2+3=0-2(x-1)^2 + 3 = 0, so (x1)2=3/2(x-1)^2 = 3/2, giving x1=±3/2x - 1 = \pm \sqrt{3/2}, i.e. x=1±6/21±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.

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