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 base graph $y = e^x$:

• Domain $\mathbb{R}$, range $(0, \infty)$. Always positive.
• $y$-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 $a > 0$, $a \neq 1$:

• $y = a^x$ 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 $\mathbb{R}$.
• $x$-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 $e^x$

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

• $h$ shifts horizontally, $k$ shifts vertically (and changes the asymptote to $y = k$).
• $a$ stretches vertically and may reflect in the $x$-axis if $a < 0$.
• $b$ 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 $\ln x$

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

• $h$ 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).
• $a$ stretches vertically; if $a < 0$, reflects.
• $k$ shifts vertically.

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

Some specific shapes

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

Sketch a transformed exponential, stage by stage

The 2022 HSC asked for a sketch of $y = 2e^x - 4$ with its asymptote and intercepts marked; the same method handles any transformed exponential. Below, $y = 3 - e^{x - 1}$ is built from $y = e^x$ one transformation at a time, tracking how the asymptote and a key point move at each step. This is the worked Step 1 below, drawn out.

Stage 1, start from the base curve. Draw $y = e^x$: increasing, always positive, through $(0, 1)$, with the horizontal asymptote $y = 0$ (the $x$-axis itself). Every later curve is this one moved or flipped.

Stage 2, shift right by 1. The exponent $x - 1$ acts inside, so it shifts the curve right by $1$. The point $(0, 1)$ moves to $(1, 1)$. The asymptote is unaffected by a horizontal shift, so it stays at $y = 0$.

Stage 3, reflect in the x-axis. The minus sign in $-e^{x-1}$ acts outside, flipping the curve top-to-bottom. It is now decreasing and lies below the $x$-axis, passing through $(1, -1)$. The asymptote is still $y = 0$, but the curve now approaches it from below.

Stage 4, shift up by 3. Adding $3$ acts outside, lifting the whole curve up by $3$. The asymptote rises with it from $y = 0$ to $y = 3$. Now read off the intercepts: the $y$-intercept is $3 - e^{-1} \approx 2.63$, and the $x$-intercept solves $e^{x-1} = 3$, giving $x = 1 + \ln 3 \approx 2.10$. The finished curve $y = 3 - e^{x-1}$ is decreasing, sits below $y = 3$, and crosses both axes.

How exam questions ask about exponential and logarithmic graphs

The phrasings recur; map each to the move:

• "Sketch $y = a e^{b(x-h)} + k$, marking the asymptote, the $y$-intercept and any $x$-intercept." Asymptote is $y = k$; $y$-intercept from $x = 0$; $x$-intercept from setting $y = 0$ and solving for $x$ with a logarithm. Mark all three (2022 HSC Q16).
• "State the domain of $y = a\ln(b(x-h)) + k$" or "find the vertical asymptote." Solve "argument $> 0$" for the domain; the asymptote is the boundary value of $x$ where the argument is zero.
• "Explain why $y = e^x$ and $y = \ln x$ are reflections in $y = x$." State that $\ln$ is the inverse of $e^x$, and an inverse graph is the reflection of the original in $y = x$ (2021 HSC Q17).
• "Find the equation of the inverse of [an exponential]" or "[a logarithm]." Swap $x$ and $y$, then isolate using $\ln$ or $e$; the asymptote and domain swap type (horizontal becomes vertical).
• "Solve $e^x = \ldots$ or $\ln x = \ldots$ from a graph", or "find where the curve cuts an axis." Use $\ln$ to undo $e$, and $e$ to undo $\ln$; give exact values like $\ln 2$ unless a decimal is asked for.
• "Sketch $y = e^{-x}$ / $-e^x$ / $\ln(-x)$ / $\ln|x|$." Each is a single reflection of the base curve; name the axis of reflection and the new domain or asymptote.