How do you graph #f(x)=(x^2-2x-8)/(x^2-9)# using holes, vertical and horizontal asymptotes, x and y intercepts?
1 Answer
see explanation.
Explanation:
#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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
solve :
#x^2-9=0rArrx^2=9rArrx=+-3#
#rArrx=-3" and " x=3" are the asymptotes"# Horizontal asymptotes occur as
#lim_(xto+-oo),f(x)toc" ( a constant)"# divide numerator/denominator by the highest power of x, that is
#x^2#
#f(x)=(x^2/x^2-(2x)/x^2-8/x^2)/(x^2/x^2-9/x^2)=(1-2/x-8/x^2)/(1-9/x^2)# as
#xto+-oo,f(x)to(1-0-0)/(1-0)#
#rArry=1" is the asymptote"# Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here, hence there are no holes.
#color(blue)"Intercepts"#
#x=0toy=(-8)/(-9)=8/9#
#rArr"y-intercept at " (0,8/9)#
#y=0tox^2-2x-8=0to(x-4)(x+2)=0#
#rArr"x-intercepts at "(-2,0)" and " (4,0)#
graph{(x^2-2x-8)/(x^2-9) [-10, 10, -5, 5]}
