How do you find the vertical, horizontal and slant asymptotes of: #(x^2-25)/(x^2+5x)#?
1 Answer
vertical asymptote:
horizontal asymptote:
slant asymptote: does not exist
Explanation:
Finding the Vertical Asymptote
Given,
#f(x)=(x^2-25)/(x^2+5x)#
Factor the numerator.
#f(x)=((x+5)(x-5))/(x(x+5)#
Cancel out any factors that appear in the numerator and denominator.
#f(x)=(x-5)/x#
Set the denominator equal to
#color(green)(|bar(ul(color(white)(a/a)color(black)(x=0)color(white)(a/a)|)))#
Finding the Horizontal Asymptote
Given,
#f(x)=(color(darkorange)1x^2-25)/(color(purple)1x^2+5x)#
Divide the
#f(x)=color(darkorange)1/color(purple)1#
#color(green)(|bar(ul(color(white)(a/a)color(black)(f(x)=1)color(white)(a/a)|)))#
Finding the Slant Asymptote
Given,
#f(x)=(x^2-25)/(x^2+5x)#
There would be a slant asymptote if the degree of the leading term in the numerator was
#:.# , the slant asymptote does not exist.