Exponential and logarithmic graphs: key features, transformations and inverse relationship

Start with $y = e^x$ (asymptote $y = 0$, through $(0, 1)$, increasing).

Vertical stretch by $2$: $y = 2 e^x$ (asymptote $y = 0$, through $(0, 2)$).

Vertical shift down by $4$: $y = 2 e^x - 4$ (asymptote $y = -4$, through $(0, -2)$).

$x$-intercept: $2 e^x = 4$, so $e^x = 2$ and $x = \ln 2 \approx 0.693$.

$y$-intercept: $(0, -2)$.

Markers reward the correct asymptote at $y = -4$, the $y$-intercept at $(0, -2)$, the $x$-intercept at $x = \ln 2$, and a smooth increasing curve.

$\ln$ is the inverse function of $e^x$, so $\ln(e^x) = x$ and $e^{\ln x} = x$ (the second only for $x > 0$).

The graph of an inverse function is the reflection of the original in $y = x$, swapping the $x$ and $y$ axes.

$y = e^x$ has horizontal asymptote $y = 0$ (as $x \to -\infty$), domain $\mathbb{R}$, range $(0, \infty)$.

$y = \ln x$ has vertical asymptote $x = 0$ (as $x \to 0^+$), domain $(0, \infty)$, range $\mathbb{R}$.

Markers expect the inverse-function statement, the reflection-in-$y = x$ fact, and accurate asymptotes and domains.

Read the transformation. The curve is $y = e^x$ shifted up by $3$; the shape and domain are unchanged, but every $y$-value is $3$ larger.

Domain. An exponential is defined for every real input, so the domain is all real $x$:

Asymptote and range. As $x \to -\infty$, $e^x \to 0$, so $y \to 3$: the horizontal asymptote is $y = 3$. Since $e^x > 0$ always, $y = e^x + 3 > 3$, giving

Marker's note: one mark for the domain $\mathbb{R}$, one for the asymptote $y = 3$ with range $(3, \infty)$. Writing the range as $[3, \infty)$ (including $3$) is the common slip; the curve approaches $y = 3$ but never reaches it.

$x$-intercept: set $y = 0$. The log is zero when its argument is $1$:

So the $x$-intercept is $(3, 0)$.

Vertical asymptote: the argument is zero at the domain edge.

The vertical asymptote is $x = 2$, and the domain is $x > 2$.

Marker's note: one mark for the intercept $(3, 0)$ from $\ln 1 = 0$, one for the asymptote $x = 2$. Solving $\ln(x-2)=0$ as $x - 2 = 0$ (confusing $\ln 1 = 0$ with argument $= 0$) is the trap.

Swap $x$ and $y$. The inverse reflects the graph in $y = x$, which algebraically swaps the variables:

Solve for $y$ using the natural log. Take $\ln$ of both sides:

Domain of the inverse. The original $y = e^{2x}$ has range $(0, \infty)$, and the domain of the inverse equals the range of the original:

Marker's note: one mark for swapping to $x = e^{2y}$, one for isolating $y = \tfrac{1}{2}\ln x$, one for the domain $(0, \infty)$. Forgetting the factor $\tfrac{1}{2}$ (writing $y = \ln x$) drops the second mark.

Describe the transformation. Adding $2$ acts outside the log, so $y = \ln x + 2$ is $y = \ln x$ shifted up by $2$. Both curves share the vertical asymptote $x = 0$ and the domain $(0, \infty)$; the shifted curve simply sits two units higher at every point.

$x$-intercept of $y = \ln x + 2$: set $y = 0$.

So the intercept is $(e^{-2}, 0)$, which lies to the left of the original curve's intercept at $(1, 0)$, consistent with lifting the curve up.

Marker's note: one mark for naming the vertical shift up by $2$, one for setting $y = 0$, one for the exact intercept $x = e^{-2}$. Leaving the answer as a decimal without the exact $e^{-2}$ form, or giving $x = -2$, loses the final mark.

Use the first point to find $A$. Substitute $(0, 5)$:

Use the second point to find $k$. Substitute $(2, 45)$ with $A = 5$:

Take the natural log of both sides:

using $\ln 9 = \ln 3^2 = 2\ln 3$.

Asymptote. With no vertical shift, $y = 5 e^{kx} \to 0$ as $x \to -\infty$, so the horizontal asymptote is $y = 0$.

Marker's note: one mark for $A = 5$, one for reaching $e^{2k} = 9$, one for $k = \ln 3$ (accept $\tfrac{1}{2}\ln 9$), one for the asymptote $y = 0$. Not simplifying $\tfrac{1}{2}\ln 9$ to $\ln 3$ is fine; writing $k = \ln 9$ (forgetting to halve) is the error.

Part (a): read the level-off line and the start value. The curve flattens towards $T = 20$, so the horizontal asymptote is

At $t = 0$ the graph starts at $85$, and $T(0) = 20 + A e^{0} = 20 + A$, so

Part (b): substitute the point $(10, 45)$. With $A = 65$:

Take the natural log:

Part (c): set $T = 30$ and solve for $t$. Using $A = 65$:

Take the natural log, then divide by $-k$:

Substituting $k = \tfrac{1}{10}\ln\tfrac{13}{5}$ gives the exact time

Marker's note: one mark for the asymptote $T = 20$ and $A = 65$ in (a); one for reaching $e^{-10k} = \tfrac{5}{13}$ and one for $k = \tfrac{1}{10}\ln\tfrac{13}{5}$ in (b); one for setting $T = 30$ to reach $e^{-kt} = \tfrac{2}{13}$ and one for the exact $t = \tfrac{10\ln(13/2)}{\ln(13/5)}$ in (c). Substituting a rounded $k$ before the final line, when an exact answer is asked, costs the last mark.

Part (a): domain and vertical asymptote. The log needs a positive argument:

The boundary where the argument is zero gives the vertical asymptote $x = -3$.

Part (b): axis intercepts. For the $y$-intercept, set $x = 0$:

For the $x$-intercept, set $f(x) = 0$, so the argument must equal $1$:

Part (c): find and confirm the inverse. Write $y = \ln(2x + 6)$ and swap $x$ and $y$:

Exponentiate both sides to undo the log:

Hence $f^{-1}(x) = \tfrac{1}{2}e^{x} - 3$, as required. As $x \to -\infty$, $e^{x} \to 0$, so $f^{-1}(x) \to -3$: the horizontal asymptote is $y = -3$. This matches the inverse relationship, since the vertical asymptote $x = -3$ of $f$ becomes the horizontal asymptote $y = -3$ of $f^{-1}$.

Marker's note: one mark for the domain and asymptote in (a); one for each intercept $(0, \ln 6)$ and $(-\tfrac{5}{2}, 0)$ in (b); one for the algebra swapping and exponentiating to the shown inverse, and one for the asymptote $y = -3$ with the domain-range-swap justification in (c). Skipping the "swap then exponentiate" steps and simply asserting the inverse forfeits the (c) working mark on a "show that".