Question

graph the rational function.
( h(x) = \frac{-2x + 8}{x^2 - 6x + 8} )
start by drawing the asymptotes (if there are any). then plot two points on each piece of the graph. finally, click on the graph - a - function button. be sure to plot a hollow dot wherever there is a \hole\ in the graph.

Explanation:

Step1: Factor numerator and denominator

Numerator: $-2x+8 = -2(x-4)$
Denominator: $x^2-6x+8 = (x-2)(x-4)$
Simplified function (for $x
eq4$): $h(x)=\frac{-2}{x-2}$

Step2: Identify holes

Common factor: $x-4$, so hole at $x=4$.
Calculate $y$-value of hole: Substitute $x=4$ into simplified function:
$h(4)=\frac{-2}{4-2}=-1$
Hole at $(4, -1)$

Step3: Find vertical asymptote

Set simplified denominator to 0: $x-2=0 \implies x=2$

Step4: Find horizontal asymptote

Degree of numerator < degree of denominator, so $y=0$

Step5: Plot points for left of $x=2$

Choose $x=0$: $h(0)=\frac{-2}{0-2}=1$ → Point $(0,1)$
Choose $x=1$: $h(1)=\frac{-2}{1-2}=2$ → Point $(1,2)$

Step6: Plot points for right of $x=2$ (excluding hole)

Choose $x=3$: $h(3)=\frac{-2}{3-2}=-2$ → Point $(3,-2)$
Choose $x=5$: $h(5)=\frac{-2}{5-2}=-\frac{2}{3}\approx-0.67$ → Point $(5, -\frac{2}{3})$

Answer:

1. Draw vertical dashed line $x=2$ (vertical asymptote)
2. Draw horizontal dashed line $y=0$ (horizontal asymptote)
3. Plot hollow dot at $(4, -1)$ (hole)
4. Plot solid points $(0,1)$, $(1,2)$ (left of $x=2$) and $(3,-2)$, $(5, -\frac{2}{3})$ (right of $x=2$)
5. Draw the two branches of the hyperbola $h(x)=\frac{-2}{x-2}$ passing through the points, approaching the asymptotes, with a hollow dot at $(4,-1)$