Question

How do you graph `f(x)=(x+3)/((x+1)(x-3))` 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+1)(x-3)=0rArrx=-1" and " x=3`

`rArrx=-1" and " x=3" are the asymptotes"`

Horizontal asymptotes occur as

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

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

`f(x)=(x/x^2+3/x^2)/(x^2/x^2-(2x)/x^2-3/x^2)=(1/x+3/x^2)/(1-2/x-3/x^2`

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

`rArry=0" 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=3/(-3)=-1larrcolor(red)" y-intercept"`

`y=0tox+3=0tox=-3larrcolor(red)" x-intercept"`
graph{(x+3)/(x^2-2x-3) [-10, 10, -5, 5]}