The function f defined by f(x) = (a)/(x+p) + q has range y != 2, axis of symmetry y = x + 1 (positive gradient), and passes through (0, -4). Find a, p, q.

For a function of the form $f(x) = \frac{a}{x+p} + q$, the horizontal asymptote is $y=q$. The range is given as $y \in \mathbb{R}, y \neq 2$.
Therefore, $q = 2$.

The axes of symmetry for a hyperbola $f(x) = \frac{a}{x+p} + q$ are $y - q = \pm (x + p)$.
We know $q=2$, so the axes are $y - 2 = \pm (x + p)$.
The axis of symmetry with a positive gradient is $y - 2 = x + p$.
Rearranging this gives $y = x + p + 2$.
We are given that this axis of symmetry is $y = x + 1$.
Comparing the constant terms:

Now the function is $f(x) = \frac{a}{x-1} + 2$.
The graph passes through the point $(0; -4)$. Substitute these coordinates into the function:

The values are $\boxed{\text{}a = 6 \text{ and } p = -1\text{}}$

From $q=2$, the horizontal asymptote is $y=2$.
From $p=-1$, the vertical asymptote is $x = -p = -(-1) = 1$.

The function is $f(x) = \frac{6}{x-1} + 2$.

• Y-intercept (set $x=0$): This is given as $(0, -4)$. $f(0) = \frac{6}{0-1} + 2 = -6 + 2 = -4$. So the y-intercept is $(0, -4)$.
• X-intercept (set $y=0$): $$0 = \frac{6}{x-1} + 2$$ $$-2 = \frac{6}{x-1}$$ $$-2(x-1) = 6$$ $$-2x + 2 = 6$$ $$-2x = 4$$ $$x = -2$$ So the x-intercept is $(-2, 0)$.

Draw the Cartesian plane.
Draw the vertical asymptote $x=1$ as a dashed line.
Draw the horizontal asymptote $y=2$ as a dashed line.
Plot the x-intercept $(-2, 0)$.
Plot the y-intercept $(0, -4)$.
Since $a=6$ is positive, the branches of the hyperbola will be in the first and third quadrants relative to the asymptotes.
Sketch the curve passing through the intercepts and approaching the asymptotes.
For a point in the first quadrant relative to the asymptotes, let $x=2$:
$f(2) = \frac{6}{2-1} + 2 = 6 + 2 = 8$. Plot $(2, 8)$.

graph TD
    A[Start] --> B{Draw Axes};
    B --> C{Draw Vertical Asymptote x=1};
    C --> D{Draw Horizontal Asymptote y=2};
    D --> E{Plot X-intercept (-2, 0)};
    E --> F{Plot Y-intercept (0, -4)};
    F --> G{Plot additional point (2, 8)};
    G --> H{Sketch Hyperbola Branch 1 (through (-2,0) and (0,-4))};
    H --> I{Sketch Hyperbola Branch 2 (through (2,8))};
    I --> J[End];

The graph should look like this:
(Due to text-based limitations, I cannot draw the graph here. Please sketch it on paper following the steps above.)