Question

$f(x)=\frac{3x^2 + 4x - 3}{x + 1}$ 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: $x+1=0 \implies x=-1$

Step2: Find slant asymptote

Divide numerator by denominator:
$$\frac{3x^2+4x-3}{x+1}=3x+1-\frac{4}{x+1}$$
Slant asymptote: $y=3x+1$

Step3: Plot points left of $x=-1$

Choose $x=-2$: $f(-2)=\frac{3(-2)^2+4(-2)-3}{-2+1}=\frac{12-8-3}{-1}=-1$
Choose $x=-3$: $f(-3)=\frac{3(-3)^2+4(-3)-3}{-3+1}=\frac{27-12-3}{-2}=-6$
Points: $(-2,-1), (-3,-6)$

Step4: Plot points right of $x=-1$

Choose $x=0$: $f(0)=\frac{3(0)^2+4(0)-3}{0+1}=-3$
Choose $x=1$: $f(1)=\frac{3(1)^2+4(1)-3}{1+1}=\frac{3+4-3}{2}=2$
Points: $(0,-3), (1,2)$

Answer:

1. Vertical Asymptote: $x=-1$ (dashed vertical line)
2. Slant Asymptote: $y=3x+1$ (dashed line)
3. Points on left segment: $(-3, -6)$, $(-2, -1)$
4. Points on right segment: $(0, -3)$, $(1, 2)$

(Graph connects the points on each side of $x=-1$, approaching the asymptotes)