Question

graph the rational function.
$f(x) = \frac{-4x^2 - 6x + 15}{6x + 9}$
start by drawing the asymptotes. then plot two points on each piece of the graph. finally, click on the graph - a - function button.

Explanation:

Step1: Find vertical asymptote

Set denominator to 0: $6x + 9 = 0$
Solve for $x$: $6x = -9 \implies x = -\frac{3}{2} = -1.5$

Step2: Find horizontal asymptote

Degree of numerator (2) > degree of denominator (1), so no horizontal asymptote. Find slant asymptote via polynomial division:
Divide $-4x^2 -6x +15$ by $6x+9$:
$$\frac{-4x^2 -6x +15}{6x+9} = -\frac{2}{3}x + \frac{15}{6x+9}$$
As $x\to\pm\infty$, $\frac{15}{6x+9}\to0$, so slant asymptote is $y = -\frac{2}{3}x$

Step3: Plot points left of $x=-1.5$

Choose $x=-3$: $f(-3)=\frac{-4(-3)^2-6(-3)+15}{6(-3)+9}=\frac{-36+18+15}{-18+9}=\frac{-3}{-9}=\frac{1}{3}\approx0.33$
Choose $x=-2$: $f(-2)=\frac{-4(4)-6(-2)+15}{6(-2)+9}=\frac{-16+12+15}{-12+9}=\frac{11}{-3}\approx-3.67$
Points: $(-3, \frac{1}{3})$, $(-2, -\frac{11}{3})$

Step4: Plot points right of $x=-1.5$

Choose $x=0$: $f(0)=\frac{0-0+15}{0+9}=\frac{15}{9}=\frac{5}{3}\approx1.67$
Choose $x=3$: $f(3)=\frac{-4(9)-6(3)+15}{6(3)+9}=\frac{-36-18+15}{18+9}=\frac{-39}{27}=-\frac{13}{9}\approx-1.44$
Points: $(0, \frac{5}{3})$, $(3, -\frac{13}{9})$

Answer:

1. Asymptotes: Vertical asymptote $x=-1.5$ (dashed line), slant asymptote $y=-\frac{2}{3}x$ (dashed line)
2. Points to plot:

• Left of vertical asymptote: $(-3, \frac{1}{3})$, $(-2, -\frac{11}{3})$
• Right of vertical asymptote: $(0, \frac{5}{3})$, $(3, -\frac{13}{9})$

1. Sketch the graph curves passing through the points, approaching the asymptotes.