Sketch and interpret graphs of exponential and logarithmic functions, including transformations, and use the inverse relationship between them

What this dot point is asking

NESA wants you to recognise and sketch the graphs of $y = e^x$ and $y = \ln x$, including any transformed versions, and to use the inverse relationship between exponential and logarithmic functions to reason about their graphs.

The answer

The exponential graph

The base graph $y = e^x$:

• Domain $\mathbb{R}$, range $(0, \infty)$. Always positive.
• IMATH_6 -intercept at $(0, 1)$.
• No $x$-intercept.
• Strictly increasing for all $x$.
• Horizontal asymptote $y = 0$ as $x \to -\infty$. Grows without bound as $x \to \infty$.
• Concave up everywhere.

For a general base $a > 0$, $a \neq 1$:

• IMATH_15 has the same shape as $y = e^x$ if $a > 1$, and is decreasing with horizontal asymptote $y = 0$ as $x \to \infty$ if $0 < a < 1$.

The logarithmic graph

The base graph $y = \ln x$:

• Domain $(0, \infty)$, range $\mathbb{R}$.
• IMATH_24 -intercept at $(1, 0)$.
• No $y$-intercept (vertical asymptote there).
• Strictly increasing for all $x > 0$.
• Vertical asymptote $x = 0$ as $x \to 0^+$. Grows without bound as $x \to \infty$ (slowly).
• Concave down everywhere.

The inverse relationship

$\ln$ is the inverse of $e^x$ on its domain:

Graphically, the graph of an inverse is the reflection of the original in the line $y = x$. Swap the coordinates of every point: $(0, 1)$ on $e^x$ maps to $(1, 0)$ on $\ln x$. The horizontal asymptote $y = 0$ on $e^x$ becomes the vertical asymptote $x = 0$ on $\ln x$. Domain and range swap.

Transformations of IMATH_42

Apply the general transformation rules. For $y = a e^{b(x - h)} + k$:

• IMATH_44 shifts horizontally, $k$ shifts vertically (and changes the asymptote to $y = k$).
• IMATH_47 stretches vertically and may reflect in the $x$-axis if $a < 0$.
• IMATH_50 controls horizontal compression and may reflect in the $y$-axis if $b < 0$.

The horizontal asymptote is always $y = k$ (the level the exponential approaches in the limit).

Transformations of IMATH_54

For $y = a \ln(b(x - h)) + k$:

• IMATH_56 shifts horizontally and moves the vertical asymptote to $x = h$ (provided $b > 0$; in general the asymptote is at the value of $x$ that makes the argument zero).
• IMATH_60 stretches vertically; if $a < 0$, reflects.
• IMATH_62 shifts vertically.

The domain is restricted to where the argument is positive: $b(x - h) > 0$.

Some specific shapes

• IMATH_64 : reflection of $e^x$ in the $y$-axis. Decreasing, asymptote $y = 0$, through $(0, 1)$.
• IMATH_69 : reflection of $e^x$ in the $x$-axis. Decreasing (in absolute height it grows), asymptote $y = 0$ from below, through $(0, -1)$.
• IMATH_74 : reflection of $\ln x$ in the $y$-axis. Domain $(-\infty, 0)$, $x$-intercept at $(-1, 0)$, vertical asymptote $x = 0$.
• IMATH_81 : defined for all $x \neq 0$. Symmetric about the $y$-axis, vertical asymptote at $x = 0$, $x$-intercepts at $x = \pm 1$.

Worked examples

Locating the asymptote of a shifted exponential

Sketch $y = 3 - e^{x - 1}$.

Start with $y = e^x$, shift right by $1$: $y = e^{x - 1}$.

Reflect in the $x$-axis: $y = -e^{x - 1}$.

Shift up by $3$: $y = 3 - e^{x - 1}$.

Asymptote: $y = 3$ (as $x \to -\infty$). $y$-intercept: $y = 3 - e^{-1} \approx 2.632$. $x$-intercept: $e^{x - 1} = 3$, so $x = 1 + \ln 3 \approx 2.099$. The graph is decreasing.

Domain and asymptote of a logarithmic transformation

Find the domain and vertical asymptote of $y = \ln(2 x - 6) + 1$.

Argument positive: $2 x - 6 > 0$, so $x > 3$. Domain $(3, \infty)$.

Vertical asymptote at $x = 3$. $x$-intercept: $\ln(2 x - 6) = -1$, so $2 x - 6 = e^{-1}$ and $x = 3 + \frac{1}{2 e} \approx 3.184$.

Using inverse to find a graph

The inverse of $y = e^{x - 1} + 2$ is found by swapping $x$ and $y$ and solving: $x = e^{y - 1} + 2$ gives $y - 1 = \ln(x - 2)$, so $y = 1 + \ln(x - 2)$. Domain $(2, \infty)$, vertical asymptote $x = 2$. As expected, the asymptote of the original ($y = 2$) becomes the vertical asymptote of the inverse ($x = 2$).

Common traps

Wrong asymptote on a shifted exponential. $y = a e^x + k$ has asymptote $y = k$, not $y = 0$. The horizontal asymptote moves with the vertical shift.

Missing domain restriction on $\ln$. $\ln(\text{negative}) =$ undefined. Always solve "argument $> 0$" before sketching.

Confusing $\ln x$ with $\log_{10} x$. In Maths Advanced, $\ln$ means natural log (base $e$) and $\log$ usually also means base $10$. Be careful when reading the question.

Forgetting $e^x$ never reaches $0$. $y = e^x > 0$ for every real $x$. The asymptote $y = 0$ is approached, not crossed.

Sketching $\ln x$ as a power curve. $\ln$ grows slowly, slower than any positive power of $x$. It also has unbounded negative values near $x = 0$, not a finite minimum.

In one sentence

$y = e^x$ has asymptote $y = 0$ and grows from positive values; $y = \ln x$ is its reflection in $y = x$ with vertical asymptote $x = 0$ and domain $(0, \infty)$; transformations follow the general rules, and the inverse relationship lets you sketch one from the other.