(b) Given $$ f(x)=\frac{-4}{x-p}+q $$ (1) If the asymptotes of $$ f $$ intersect at the point (5;1), determine the values of $$ p $$ and $$ q $$. (2) $$ \begin{array}{l} ext { Determine the equation of the line of symmetry of } f(x) ext { which has a } \ ext { negative gradient. }\end{array} $$

Question

(b) Given $$ f(x)=\frac{-4}{x-p}+q $$ (1) If the asymptotes of $$ f $$ intersect at the point (5;1), determine the values of $$ p $$ and $$ q $$. (2) $$ \begin{array}{l}	ext { Determine the equation of the line of symmetry of } f(x) 	ext { which has a } \ 	ext { negative gradient. }\end{array} $$

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To analyze the function $$ f(x) = \frac{-4}{x - p} + q $$, we first need to identify the vertical and horizontal asymptotes.

### (1) Finding $$ p $$ and $$ q $$

**Vertical Asymptote:**
The vertical asymptote occurs where the denominator is zero, which gives us:
$$
x - p = 0 \implies x = p
$$

**Horizontal Asymptote:**
As $$ x $$ approaches infinity, the term $$ \frac{-4}{x - p} $$ approaches 0. Therefore, the horizontal asymptote is:
$$
y = q
$$

Given that the asymptotes intersect at the point (5, 1), we have:
$$
p = 5 \quad \text{(from the vertical asymptote)}
$$
$$
q = 1 \quad \text{(from the horizontal asymptote)}
$$

Thus, the values of $$ p $$ and $$ q $$ are:
$$
\boxed{p = 5, \, q = 1}
$$

### (2) Equation of the Line of Symmetry

The function $$ f(x) = \frac{-4}{x - 5} + 1 $$ can be rewritten as:
$$
f(x) = -\frac{4}{x - 5} + 1
$$

The line of symmetry for a rational function of this form is typically vertical and is located at the vertical asymptote. However, since we are asked for a line of symmetry with a negative gradient, we need to find the line that reflects the function across a certain point.

The function $$ f(x) $$ is symmetric about the vertical line $$ x = p $$ (which is $$ x = 5 $$). To find a line of symmetry with a negative gradient, we can consider a line that passes through the point of intersection of the asymptotes (5, 1) and has a negative slope.

Let’s denote the line of symmetry as:
$$
y - 1 = m(x - 5)
$$
where $$ m < 0 $$ (negative gradient).

To find a specific line, we can choose a slope. For example, let’s take $$ m = -1 $$:
$$
y - 1 = -1(x - 5)
$$
$$
y - 1 = -x + 5
$$
$$
y = -x + 6
$$

Thus, the equation of the line of symmetry with a negative gradient is:
$$
\boxed{y = -x + 6}
$$
(1) $$ p = 5, \, q = 1 $$ (2) $$ y = -x + 6 $$

(b) Given
(1) If the asymptotes of intersect at the point (5;1), determine the
values of and .
(2)

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(b) Given (1) If the asymptotes of intersect at the point (5;1), determine the values of and . (2)