How do you graph #y=(2x^2-9x-5)/(x^2-16)# using asymptotes, intercepts, end behavior?

1 Answer
Oct 20, 2016

Answer:

The vertical asymptotes are #x=4# and #x=-4#
And the horizontal asymptote is #y=2#

Explanation:

Factorising numerator and denominator
#y=((2x+1)(x-5))/((x+4)(x-4))#
To factorise you can use the formula#x=(-b+-(b^2-4ac))/(2a)#
We can see that #x!=+-4# as we cannot divide by 0. Hence we obtain the vertical asymptotes #x=4# and #x=-4#
And the limit of y as #x->oo# is 2. therefore the horizontal asymptote is #y=2#
When #x=0# , #y=5/16#
graph{(2x^2-9x-5)/(x^2-16) [-20, 20, -10, 10]}