Question

draw a graph of $f(x) = \frac{x - 9}{-x + 3}$ by first placing the horizontal and vertical asymptotes, then plotting an additional point on the graph.

Explanation:

Step1: Find vertical asymptote

Set denominator to 0:
$-x + 3 = 0 \implies x = 3$

Step2: Find horizontal asymptote

Compare leading degrees:
$\lim_{x \to \infty} \frac{x - 9}{-x + 3} = \lim_{x \to \infty} \frac{1 - \frac{9}{x}}{-1 + \frac{3}{x}} = -1$

Step3: Plot an additional point

Choose $x=0$:
$f(0) = \frac{0 - 9}{-0 + 3} = \frac{-9}{3} = -3$
So point is $(0, -3)$

Answer:

1. Vertical asymptote: $x=3$ (dashed vertical line at $x=3$)
2. Horizontal asymptote: $y=-1$ (dashed horizontal line at $y=-1$)
3. Plot the point $(0, -3)$, then sketch the hyperbola approaching both asymptotes, with one branch passing through $(0, -3)$ and the other on the opposite side of the vertical asymptote.