Logarithmic graphs, domains and transformations | Year 12

01 / LEARN

The idea, step by step

Start with the basic graph

For $y = ln x$ , the domain is $x > 0$ , the range is all real numbers and the vertical asymptote is $x = 0$ . Useful points include $(1, 0)$ , $(e, 1)$ and $(1 / e, - 1)$ . The graph is increasing and concave down. A vertical asymptote is approached, not crossed by this graph.

Read the input before drawing

For $y = a ln (b (x - h)) + k$ , where $a \neq 0$ and $b \neq 0$ , first solve $b (x - h) > 0$ . The asymptote is $x = h$ , with the graph on the side selected by the sign of $b$ . The coefficient $a$ stretches vertically and reflects in the x-axis if negative; $k$ shifts vertically. Transform convenient points or solve for intercepts instead of guessing from a small calculator window.

A common trap

The domain of $ln (5 - x)$ is $x > 5$ .

Instead: Solve the input inequality $5 - x > 0$ . It gives $x < 5$ , so the graph lies to the left of its asymptote.

02 / WORKED EXAMPLES

See the thinking

Example 1

Translate a graph

Give the domain, asymptote and intercepts of $y = ln (x - 1) + 1$ .

The input condition is $x - 1 > 0$ , so the domain is $x > 1$ and the asymptote is $x = 1$ .
For an x-intercept, solve $ln (x - 1) = - 1$ , giving $x = 1 + e^{- 1}$ .
The graph has no y-intercept because $x = 0$ is outside the domain.

Domain

(1, \infty)

; asymptote

x = 1

; x-intercept

(1 + e^{- 1}, 0)

; no y-intercept.

Example 2

Reflect and restrict

For $g (x) = 2 - ln (4 - x)$ , find the domain and the point where $g (x) = 2$ .

$4 - x > 0$ gives $x < 4$ ; the vertical asymptote is $x = 4$ .
$2 - ln (4 - x) = 2$ gives $ln (4 - x) = 0$ .
Thus $4 - x = 1$ , so $x = 3$ .

Domain

(- \infty, 4)

and point

(3, 2)

Check your understanding

What is the vertical asymptote of $y = 3 ln (x + 2) - 5$ ?

Show answer

$x = - 2$

The logarithm input tends to zero at $x = - 2$ ; its positive-input domain is $x > - 2$ .

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

State the domain and asymptote of $y = {log}_{2} (x - 3)$ .
2 marks
Find the x-intercept of $y = ln x - 2$ .
2 marks
Find the domain and y-intercept of $y = ln (6 - 2 x)$ .
3 marks
Describe the transformation from $y = ln x$ to $y = - 2 ln x + 3$ .
2 marks
Find the inverse of $f (x) = ln (x + 2) - 1$ and its domain.
3 marks
Sketch $y = ln x$ and $y = ln (2 x)$ on the same axes using exact anchor points. State their vertical separation.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

State the domain and asymptote of $y = {log}_{2} (x - 3)$ .
2 marks
The input $x - 3$ must be positive; its zero gives the asymptote.
$x > 3$ ; $x = 3$
Find the x-intercept of $y = ln x - 2$ .
2 marks
$ln x = 2$ gives $x = e^{2}$ .
$(e^{2}, 0)$
Find the domain and y-intercept of $y = ln (6 - 2 x)$ .
3 marks
$6 - 2 x > 0$ gives $x < 3$ ; substitute $x = 0$ .
$x < 3$ ; $(0, ln 6)$
Describe the transformation from $y = ln x$ to $y = - 2 ln x + 3$ .
2 marks
The multiplier $- 2$ changes output values; the $+ 3$ is applied afterwards.
Reflection in the x-axis, vertical scale factor 2, then translation up 3.
Find the inverse of $f (x) = ln (x + 2) - 1$ and its domain.
3 marks
$y + 1 = ln (x + 2)$ gives $x = e^{y + 1} - 2$ . The original range is all real numbers.
$f^{- 1} (x) = e^{x + 1} - 2$ , $x \in ℝ$
Sketch $y = ln x$ and $y = ln (2 x)$ on the same axes using exact anchor points. State their vertical separation.
3 marks
For $x > 0$ , $ln (2 x) = ln x + ln 2$ . Anchor points include $(1, 0)$ and $(1, ln 2)$ .
The second graph is $ln 2$ units above the first; both have asymptote $x = 0$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: logarithmic functions and their applications; reciprocal-linear antiderivatives are taught in the Integration chapter.

Official curriculum source ↗

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