Question

How do you graph `f(x)=(-3x^2-12x-9)/(x^2+5x+4)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

see explanation.

Explanation:

First, factorise and simplify f(x)

`f(x)=(-3cancel((x+1))(x+3))/(cancel((x+1))(x+4))=(-3(x+3))/(x+4)`

with exclusion x ≠ - 1 which indicates a hole at x = - 1

The graph of `f(x)=(-3(x+3))/(x+4)` is the same as

`(-3x^2-12x-9)/(x^2+5x+4)` but without the hole.

`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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve : `x+4=0rArrx=-4" is the asymptote"`

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`f(x)=((-3x)/x-9/x)/(x/x+4/x)=(-3-9/x)/(1+4/x)`

as `xto+-oo,f(x)to(-3-0)/(1+0)`

`rArry=-3" is the asymptote"`

`color(blue)"Approaches to asymptotes"`

`"horizontal asymptote " y=-3`

as `xto+oo,f(x)toy=-3" from above"`

as `xto-oo,f(x)toy=-3" from below"`

`"vertical asymptote "x=-4`

`lim_(xto-4^-)f(x)to-oo`

`lim_(xto-4^+)f(x)to+oo`

`color(blue)"Intercepts"`

`x=0toy=-9/4larr" y-intercept"`

`y=0tox+3=0tox=-3larr" x-intercept"`
graph{(-3(x+3))/(x+4) [-10, 10, -5, 5]}