Solution:
Given:
[tex]y=\frac{9x^2+81x}{x^3+8x^2-9x}[/tex]
The graph of the function is shown;
1) x-intercepts
From the graph of the equation, there is no x-intercept because the curve does not cut the x-axis at any point.
2) Vertical asymptote
The vertical asymptote is the vertical line the curve tends towards but does not touch.
From the graph, the vertical asymptote is at x = 1.
3) Horizontal asymptote
The horizontal asymptote is the horizontal line the curve tends towards but does not touch.
From the graph, the horizontal asymptote is at y = 0.
4) Hole
To get the hole, we get the common factor.
[tex]\begin{gathered} y=\frac{9x^2+81x}{x^3+8x^2-9x} \\ y=\frac{9x\left(x+9\right)}{x\left(x-1\right)\left(x+9\right)} \\ The\text{ common factors are;} \\ x\text{ and }x+9 \\ \\ Equate\text{ the common factors to zero} \\ Hence,\text{ there is a hole at}x=0 \\ \\ \\ \\ x+9=0 \\ Also,\text{ there is a hole at}x=-9 \\ \\ \\ Hence,\text{ cancelling out the common terms, the equation becomes;} \\ y=\frac{9}{(x-1)} \\ when\text{ x = 0,} \\ y=\frac{9}{0-1}=\frac{9}{-1} \\ y=-9 \\ Hence,\text{ hole is at }(0,-9) \\ \\ \\ Also,\text{ when x = -9} \\ y=\frac{9}{-9-1}=\frac{9}{-10} \\ Hence,\text{ hole is also at }(-9,-\frac{9}{10}) \end{gathered}[/tex]
Therefore, there are two holes.
A hole exists at (0,-9) which is where the y-intercept should be because the y-intercept usually exists at the point x = 0. Hence, there is a hole where the y-intercept should be. TRUE
