Given the function $p(x) = igg(\frac{1}{3}\bigg)^x$ - NSC Mathematics - Question 4 - 2023 - Paper 1

To find the inverse, we swap $x$ and $y$:

1. Let $y = \left(\frac{1}{3}\right)^x$.

2. Taking the logarithm:

   $x = \left(\frac{1}{3}\right)^y$

   Taking the logarithm gives us $y = - \log_{\frac{1}{3}}(x)$, or equivalently,

   $y = \log_3\left(\frac{1}{x}\right)$.

The domain of $p^{-1}$ is all real numbers, $\mathbb{R}$, since the inverse function is defined for all positive values of $x$.

The vertical asymptote of $p(x)$ is given by the line $y = 5$.

The vertical asymptote occurs at $x = 1$, and the horizontal asymptote occurs at $y = 2$.

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

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

1. This simplifies to: $\frac{4}{x - 1} = -2$

2. Therefore, solving gives: $4 = -2(x - 1)$ $4 = -2x + 2$ $2x = -2 \implies x = -1$.

The sketch should show the horizontal asymptote at $y = 2$ and the vertical asymptote at $x = 1$. The x-intercept is at $(-1, 0)$, and the y-intercept can be found by substituting $x = 0$:

$f(0) = \frac{4}{0 - 1} + 2 = -4 + 2 = -2$

Thus, the y-intercept is at $(0, -2)$.

This simplifies to:

$\frac{4}{x - 1} = 0\implies x \neq 1$,

so the solution to this under the constraint is all $x > 1$.

To find the axis of symmetry that has a negative gradient near the point $(3, 2)$ which lies on $y = x + c$ form, we have:

1. Equation: $y = -x + c$,
2. Substitute the coordinates of the point: If $c = 5$, then: $y = -x + 5$.

Thus, the final equation is $y = -x + 5$.