Question

How do you graph `f(x)=(x^2-2x-8)/(x^2-9)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

see explanation.

Explanation:

`color(blue)"Asymptotes"`

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : `x^2-9=0rArrx^2=9rArrx=+-3`

`rArrx=-3" and " x=3" are the asymptotes"`

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" ( a constant)"`

divide numerator/denominator by the highest power of x, that is `x^2`

`f(x)=(x^2/x^2-(2x)/x^2-8/x^2)/(x^2/x^2-9/x^2)=(1-2/x-8/x^2)/(1-9/x^2)`

as `xto+-oo,f(x)to(1-0-0)/(1-0)`

`rArry=1" is the asymptote"`

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here, hence there are no holes.

`color(blue)"Intercepts"`

`x=0toy=(-8)/(-9)=8/9`

`rArr"y-intercept at " (0,8/9)`

`y=0tox^2-2x-8=0to(x-4)(x+2)=0`

`rArr"x-intercepts at "(-2,0)" and " (4,0)`
graph{(x^2-2x-8)/(x^2-9) [-10, 10, -5, 5]}