2019 H2 Maths Paper 1, Question 4

Since $2^{x} > 0$ for all $x$, $2^{x} - 10 \rightarrow - 10$ as $x \rightarrow - \infty$ and $2^{x} - 10 \rightarrow \infty$ as $x \rightarrow \infty$, increasing throughout, with a single zero where $2^{x} = 10$.

$y\text{-intercept}$: at $x = 0$,

$x\text{-intercept}$: where $2^{x} - 10 = 0$,

Since $2^{x} - 10 \rightarrow - 10$ as $x \rightarrow - \infty$, $\left| 2^{x} - 10 \right| \rightarrow 10$, so $y = 10$ is an asymptote on the left. For $x < \log_{2}10$, $2^{x} - 10 < 0$ so $y = 10 - 2^{x}$, decreasing from $10$ towards $0$. For $x > \log_{2}10$, $2^{x} - 10 > 0$ so $y = 2^{x} - 10$, increasing from $0$ without bound.

Since $\left| 2^{x} - 10 \right| \leq 6$,

Note: $2^{x}$ is increasing for all $x$, so applying $\log_{2}$ to a double inequality preserves both inequality signs.