The graphs of the functions $f(x) = a(x + p)^2 + q$ and $g(x) = \frac{k}{x + r} + d$ are sketched below - NSC Mathematics - Question 5 - 2016 - Paper 1

Using the information that the horizontal asymptote of $g$ is $y = -2$, we see:

$d = -2$

Then we also set up an equation with the known points:

For the vertical asymptote, substituting into $g(x)$ gives: $k = -2r$

By substituting $x = 1$ into $g(x)$, we relate:

$k = 6 - 2r$

From these we solve:

• Equating gives us two equations:
  1. $k = -2r$
  2. $k = 6 - 2r$

Solve these simultaneously to get: $k = 6, r = -3, d = -2$.

To solve for $x$ where $g(x) \geq f(x)$ for $x \leq 1$, we analyze:

• Set up the inequality: $\frac{6}{x - 3} - 2 \geq 4(x + 2)^2 - 4$

• Simplify to: $6 \geq 4(x + 2)^2 (x - 3) + 2(x - 3)$

• Solve this inequality considering the roots and the interval restrictions.

The critical points need to evaluate where this holds. Only values $x \in [-8, -4)$ satisfy the condition in our required interval.

We set: $4(k + 2)^2 - 4 = k$

This leads to: $4(k + 2)^2 - k - 4 = 0$

By examining the discriminant $\Delta$, we have:

$\Delta = b^2 - 4ac$

For two unequal roots: $\Delta > 0$

Solving for $k$, we find conditions such that $k$ must be less than a certain threshold, leading to necessary values $k < -4$.

The axis of symmetry for a rational function $g(x)$ will be in the form: $y = -x + c$

Using the intersection point $(\frac{3}{2}, -2)$ and the point's slope:

• Substituting into $y = -x + c$ gives: $-2 = -\frac{3}{2} + c \\ c = -2 + \frac{3}{2} = -\frac{1}{2}$

Thus the equation becomes: $y = -x - \frac{1}{2}$.

To reflect point $P(1, -8)$ across the line $y = -x - \frac{1}{2}$:

1. Find the perpendicular from $P$ to the line. 2. Solve for intersection coordinates and use them to compute the reflection points.

After calculations, we find: $Q = (\frac{15}{2}, \frac{3}{2})$.