How do you find the vertical, horizontal or slant asymptotes for #(x² - 3x - 7)/(x+3) #?

1 Answer
Nov 3, 2016

Answer:

The vertical asymptote is #x=3#
and the slant asymptote is #y=x-6#

Explanation:

As we cannot divide by zero, so #x!=-3#
#:. x=-3# is a vertical asymptote

As the degree of the numerator is greater than the degree of the denominator, we would expect a slant asymptote. W e have to do a long division.
#color(white)(aaaa)##x^2-3x-7##color(white)(aaaa)##∣##x+3#
#color(white)(aaaa)##x^2+3x##color(white)(aaaaaaaa)##∣##x-6#
#color(white)(aaaaa)##0-6x-7#
#color(white)(aaaaaaa)##-6x-18#
#color(white)(aaaaaaaaa)##0+11#

Finally we have, #(x^2-3x-7)/(x-3)=x-6+11/(x+3)#
The slant asymptote is #y=x-6#

graph{(y-((x^2-3x-7)/(x+3)))(y-x+6)=0 [-58.5, 58.5, -29.27, 29.28]}