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

1 Answer
Jan 22, 2017

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]}