The diagram below shows the graphs of $g(x) = \frac{2}{x + p} + q$ and $f(x) = \log_2 x.$ - $y = -1$ is the horizontal asymptote of $g$ - NSC Mathematics - Question 5 - 2017 - Paper 1

To determine the equation of $g$, we identify the factors that characterize its behavior. Given that the vertical asymptote is determined by $g(x) = \frac{2}{x + p} + q$, we have:

Setting the horizontal asymptote at $y = -1$, we can deduce that:

$q = -1$

Thus, the equation simplifies to:

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

We know that at point A, $A(t; 1)$, $g(t) = f(1) = \log_2(1) = 0$.

Setting up the equation of $g$ using the earlier form:

$0 = \frac{2}{t + p} - 1$

Solving for $t$ yields:

$1 = \frac{2}{t + p} \implies t + p = 2 \implies t = 2 - p$

Substituting known values (if given) would provide a numeric solution for $t$.

Starting from the function $f(x) = \log_2 x$, the inverse can be found by swapping $x$ and $y$:

$x = \log_2 y$

To express this in the form of $y = ...$, we exponentiate:

$y = 2^x$

First, we rewrite the inequality using the expression for the inverse:

$f^{-1}(g(x)) < 3 \implies g(x) < 2^3 = 8$

Next, substituting our earlier equation of $g$ gives us:

$\frac{2}{x + p} - 1 < 8$

This can be solved for $x$ based on parameters set by $p$.

The axis of symmetry is given by the equation of the vertical asymptote derived from $g$. In general, finding the axis of symmetry would yield the x-intercept. The x-coordinate is $t = 1$ from our earlier deductions:

Setting up the intersection condition:

$f(x) = g(x)$

This intersection must yield a negative gradient, thus ensuring that:

$\frac{dy}{dx} < 0$

For specific computations, substituting into the equations will reveal valid intersection points.