The diagram below shows the graphs of $g(x) = \frac{2}{x + p} + q$ and $f(x) = \log_{a} x$ - NSC Mathematics - Question 5 - 2017 - Paper 1

To determine the equation of $g$, we need to analyze the components provided:
Given the vertical asymptote intersects the x-axis at $E$, we can express the equation of $g$ as:

$g(x) = \frac{2}{x - 2} - 1$
This is obtained by considering the transformations reflected in the given diagram.

To find the value of $t$, we know $A(t; 1)$ is a point of intersection between $f$ and $g$. Substituting $1$ into the equation of $f(x)$ gives:

$f(t) = \log_{a} t = 1$ This implies:

$t = a^{1} = a$

We also know from the equation of $g$: $g(t) = \frac{2}{t - 2} - 1$ Setting $g(t)$ equal to $1$ allows us to solve for $t$:

1 + 1 = \frac{2}{t - 2} \\ 2(t - 2) = 2 \\ \Rightarrow t - 2 = 1 \\ \Rightarrow t = 3$$

To find the inverse of $f(x) = \log_{a} x$, we interchange $x$ and $y$:

$x = \log_{a} y$ Exponentiating both sides yields:

$y = a^{x}$ Thus, the equation of $f^{-1}$ is:

$y = a^{x}$

We need to solve:

$f^{-1}(g) < 3 \Rightarrow g < a^{3}$ Using the determined equation of $g$:

$\frac{2}{x - 2} - 1 < a^{3}$ Rearranging gives:

$\Rightarrow \frac{2}{x - 2} < a^{3} + 1$ Thus, the inequality must be solved for $x$:

$x < \frac{2}{(a^{3} + 1)} + 2$

The axis of symmetry is given by the equation: $y = -x + 1$ To find the intersection with $f(x)$:

$\log_{a} x = -x + 1$ We calculate the point of intersection: Solving this equation will provide the coordinates of the intersection that has a negative gradient.