SOLUTION: How dou you find the domain and their vertical, horizontal or oblique asymptote? r(x) = x^2+x-72/ x^2-x-56

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Question 890057: How dou you find the domain and their vertical, horizontal or oblique asymptote?
r(x) = x^2+x-72/ x^2-x-56
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
r(x) = x^2+x-72/ x^2-x-56
The correct text form you want is r(x)=(x^2+x-72)/(x^2-x-56)
and as rendered looks like r%28x%29=%28x%5E2%2Bx-72%29%2F%28x%5E2-x-56%29



Factor if you can.
r%28x%29=%28%28x-8%29%28x%2B9%29%29%2F%28%28x%2B7%29%28x-8%29%29.
A discontinuity occurs at x=8. This is because of the factor %28x-8%29%2F%28x-8%29.
A vertical asymptote will occur at x=-7 because the function is undefined there.

Domain: -infinity%3Cx%3C-7 U -7%3Cx%3C8 U 8%3Cx%3Cinfinity.

graph%28300%2C300%2C-15%2C10%2C-15%2C10%2C%28x%5E2%2Bx-72%29%2F%28x%5E2-x-56%29%29
The graph will not display properly the discontinuity at x=8 but it is there. The HORIZONTAL asymptote is y=1. This might be easier to understand if you take your original function before factoring and examine what happens for x extremely large or small without bound. The x^2 terms in numerator and denominator become increasingly far more important, so each approaches the same square value, and you have the ratio of these.