Given the function $p(x) = \left( \frac{1}{3} \right)^x$: 4.1.1 Is $p$ an increasing or decreasing function? 4.1.2 Determine $p^{-1}$, the inverse of $p$, in the form $y = ...$ 4.1.3 Write down the domain of $p^{-1}$ - NSC Mathematics - Question 4 - 2023 - Paper 1

To find the inverse of $p(x)$:

1. Start with $y = \left( \frac{1}{3} \right)^x$.
2. Swap $x$ and $y$: $x = \left( \frac{1}{3} \right)^y$.
3. Take the logarithm of both sides: $\log_{\frac{1}{3}}(x) = y$.

Thus, the inverse is $p^{-1}(x) = \log_{\frac{1}{3}}(x)$.

The domain of the inverse function $p^{-1}(x) = \log_{\frac{1}{3}}(x)$ is $x > 0$, since logarithms are defined only for positive numbers.

The asymptote of the function $p(x)$ occurs where $p(x)$ approaches a constant as $x$ approaches infinity. Thus, for $p(x) - 5$, the equation of the asymptote is $y = -5$.

The function $f(x) = \frac{4}{x - 1} + 2$ has:

• A vertical asymptote at $x = 1$ (where the denominator is zero).
• A horizontal asymptote at $y = 2$ (as $x \to \infty$).

To find the x-intercept, set $f(x) = 0$:

$0 = \frac{4}{x - 1} + 2$

Rearranging gives:

$\frac{4}{x - 1} = -2 \\ 4 = -2(x - 1) \\ 4 = -2x + 2 \\ 2 = -2x \\ x = -1$

Thus, the x-intercept is $x = -1$.

To sketch the graph of $f(x)$:

1. Mark the vertical asymptote at $x = 1$ and the horizontal asymptote at $y = 2$.
2. Plot the x-intercept at $(-1, 0)$.
3. The graph approaches the horizontal asymptote as $x$ approaches infinity and diverges as it approaches the vertical asymptote.

To find the values of $x$ for which $\frac{4}{x - 1} = -2$, we solve:

1. From our previous calculation: $x = 2$

Thus, $x = 2$ is the value satisfying the equation.

The axis of symmetry for the function $f(x - 2) = \frac{4}{(x - 2) - 1} + 2$ is determined by finding the x-coordinate of the vertex. The transformation $f(x - 2)$ shifts the function two units to the right. The symmetry line would be at $x = 2$, with a negative gradient of the original function.