How do you graph #y=(x^2-7x-60)/(x+3)# using asymptotes, intercepts, end behavior?

1 Answer
Nov 7, 2016

Answer:

See the graph below

Explanation:

As you cannot divide by #0#, #x=-3# is a vertical asymptotes.
As the degree of the numerator #># the degree of the denominator, we expect a slant asymptote. So we do a long division.
#color(white)(aaa)##x^2-7x-60##color(white)(aaaa)##∣##x+3#
#color(white)(aaa)##x^2+3x##color(white)(aaaaaaaa)##∣##x-10#
#color(white)(aaa)##0-10x-60#
#color(white)(aaaaa)##-10x-30#
#color(white)(aaaaaaaaa)##0-30#

#y=(x^2-7x-60)/(x+3)=x-10-30/(x+3)#
So #y=x-10# is an oblique asymptote
#lim_(+-oo)y=lim_(+-oo)x^2/x=+-oo#
When #x=0##=>##y=-20#
graph{(y-(x^2-7x-60)/(x+3))(y-x+10)=0 [-83.3, 83.26, -41.65, 41.7]}