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Functions and graphs: QCE Math Methods Unit 1 Year 11

Syllabus dot point

“Functions and graphs introduced in Year 11, including linear, quadratic, cubic, polynomial, exponential and logarithmic functions; their key features, intercepts and transformations”

9 min read

Quick answer

Unit 1 introduces the major function families (linear, quadratic, polynomial, exponential, logarithmic) and the four standard transformations (translation and dilation in both xx and yy); sketching requires all key features (intercepts, turning points, asymptotes, end behaviour) labelled.

Jump to a section
  1. What this dot point is asking
  2. Function families
  3. Key features to identify
  4. The four standard transformations
  5. Order of transformations
  6. Reading key features off each family
  7. Exam-style questions

What this dot point is asking

QCAA wants Year 11 students to recognise the major function families, identify their key features (intercepts, turning points, asymptotes and end behaviour), and apply the four standard transformations to sketch and describe graphs. This is the foundation for the Unit 3 and 4 calculus work, where the same families and transformations reappear with derivatives and integrals.

Function families

Linear y=mx+cy = mx + c. Straight line, gradient mm, yy-intercept cc.

Quadratic y=ax2+bx+cy = ax^2 + bx + c or vertex form y=a(x−h)2+ky = a(x - h)^2 + k. Parabola, turning point (h,k)(h, k).

Cubic y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d. Up to two turning points; can be monotonic.

Polynomial. Degree nn polynomial has up to n−1n - 1 turning points.

Exponential y=axy = a^x for a>0,a≠1a > 0, a \neq 1. Always positive, horizontal asymptote y=0y = 0.

Logarithmic y=log⁡a(x)y = \log_a(x). Defined for x>0x > 0, vertical asymptote x=0x = 0, inverse of exponential.

Key features to identify

When analysing or sketching any function, work through a fixed checklist so nothing is missed:

  • Domain and range (the set of allowed inputs and resulting outputs).
  • xx- and yy-intercepts (set y=0y = 0 and x=0x = 0 respectively).
  • Turning points or stationary points (maxima, minima, inflections).
  • Asymptotes, both vertical and horizontal, where the graph approaches but never reaches a line.
  • End behaviour, the direction the graph heads as x→±∞x \to \pm\infty.

A function is a rule assigning exactly one output to each input, which is why graphs of functions pass the vertical line test: no vertical line meets the graph more than once.

The four standard transformations

Given y=f(x)y = f(x):

Translation in yy
y=f(x)+ky = f(x) + k shifts up by kk.
Translation in xx
y=f(x−h)y = f(x - h) shifts right by hh.
Dilation in yy
y=af(x)y = af(x) stretches vertically by factor aa.
Dilation in xx
y=f(bx)y = f(bx) compresses horizontally by factor 1/b1/b.

Reflections are special cases: a<0a < 0 reflects in the xx-axis, and b<0b < 0 reflects in the yy-axis.

Order of transformations

When several transformations combine, apply the operations on xx (inside the function) before the operations on yy (outside). For y=af(b(x−h))+ky = af(b(x - h)) + k, work from the inside out: first the horizontal dilation by 1b\tfrac{1}{b}, then the horizontal translation by hh, then the vertical dilation by aa, then the vertical translation by kk. Getting this order wrong is the most common source of misplaced graphs, because a horizontal shift applied before a horizontal dilation lands in a different place.

Reading key features off each family

Each family advertises its features in its equation. A quadratic in vertex form shows its turning point directly; an exponential y=ax+ky = a^x + k has horizontal asymptote y=ky = k; a logarithm y=log⁡a(x−h)y = \log_a(x - h) has vertical asymptote x=hx = h. Identifying the family first, then reading the relevant features, is faster and more reliable than plotting points blindly.

Building a graph by transformations

Start from the parent function

Sketch y=3(x−2)3−5y = 3(x - 2)^3 - 5 by transforming y=x3y = x^3, which has its point of inflection at the origin.

Apply the inside operation

The (x−2)(x - 2) translates the graph right by 22, moving the inflection point to (2,0)(2, 0).

Apply the outside operations

The factor 33 is a vertical dilation by factor 33 (steepening the curve), and the −5-5 translates it down by 55, moving the inflection point to (2,−5)(2, -5).

State the result

The final curve is a cubic with point of inflection (2,−5)(2, -5), steeper than y=x3y = x^3, increasing throughout. Markers reward naming each transformation and tracking the inflection point through the shifts.

Sketching an exponential

Sketch y=2x−3y = 2^x - 3, labelling the asymptote and both intercepts.

Step 1: Identify the parent function and how it has been transformed

The parent is y=2xy = 2^x, which is always positive and has horizontal asymptote y=0y = 0. Subtracting 33 is a vertical translation downward by 33, which shifts the asymptote from y=0y = 0 to y=−3y = -3. The curve still increases from left to right because the base b=2>1b = 2 > 1.

Step 2: Find the yy-intercept

Substitute x=0x = 0 into the equation:

y=20−3=1−3=−2,y = 2^0 - 3 = 1 - 3 = -2,

so the yy-intercept is (0,−2)(0, -2).

Step 3: Find the xx-intercept

Set y=0y = 0 and solve for xx. We need 2x=32^x = 3, which gives:

x=log⁡23≈1.58.x = \log_2 3 \approx 1.58.

The curve crosses the xx-axis just above x=1.5x = 1.5.

Step 4: Describe the complete shape

The curve approaches y=−3y = -3 from above as x→−∞x \to -\infty (never reaching the asymptote), passes through (0,−2)(0, -2) and (log⁡23,0)(\log_2 3, 0), and increases without bound as x→+∞x \to +\infty.

Final answer: Asymptote y=−3y = -3; yy-intercept (0,−2)(0, -2); xx-intercept (log⁡23,0)≈(1.58,0)(\log_2 3, 0) \approx (1.58, 0).

Common errors
Translation sign error
f(x−h)f(x - h) shifts right by hh, not left.
Wrong transformation order
Apply inside-the-bracket first (operations on xx), then outside (operations on yy).
Forgetting asymptotes
Exponentials have horizontal asymptotes; logs have vertical asymptotes. Mark them.

Exam-style questions

Questions in the style of QCAA 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.

Practice4 marks
Paper 1 (technique). Sketch y=−2(x−3)2+8y = -2(x - 3)^2 + 8, labelling the turning point, the yy-intercept and any xx-intercepts.
Show worked answer →

Vertex form y=a(x−h)2+ky = a(x - h)^2 + k gives turning point (3,8)(3, 8), opening downward since a=−2<0a = -2 < 0.

yy-intercept: y=−2(0−3)2+8=−18+8=−10y = -2(0 - 3)^2 + 8 = -18 + 8 = -10, so (0,−10)(0, -10).

xx-intercepts: −2(x−3)2+8=0-2(x - 3)^2 + 8 = 0, so (x−3)2=4(x - 3)^2 = 4, x−3=±2x - 3 = \pm 2, giving x=1x = 1 and x=5x = 5.

Markers reward reading the turning point from vertex form, both intercepts, and the downward shape.

Practice4 marks
Paper 1 (technique). The graph of y=f(x)y = f(x) has a vertical asymptote at x=0x = 0 and passes through (1,0)(1, 0). Describe the single graph y=2f(x−3)+1y = 2f(x - 3) + 1 in terms of the transformations applied, and state the equation of its asymptote.
Show worked answer →

Apply the transformations to ff: f(x−3)f(x - 3) translates right 33; the factor 22 is a vertical dilation by factor 22; the +1+1 translates up 11.

The vertical asymptote moves with the horizontal translation, from x=0x = 0 to x=3x = 3. (The vertical dilation and upward shift do not move a vertical asymptote.)

Markers reward naming each of the three transformations in order and tracking the asymptote through the horizontal translation only.

Practise this

Sources & how we know this

ExamExplained