Asked by Aaron
Determine the holes, vertical asymptotes and horizontal asymptotes of the rational function y=(3x^(2)+8x-10)/(x^(2)+7x+12)
Hole:
Vertical Asymptote:
Horizontal Asymptote:
Hole:
Vertical Asymptote:
Horizontal Asymptote:
Answers
Answered by
Steve
y = (3x^2+8x-10)/(x^2+7x+12)
= (3x^2+8x-10) / (x+3)(x+4)
no holes, since y is never 0/0
vertical asymptotes where the denominator is zero: x = -3,-4
as x gets huge, y -> 3x^2/x^2 = 3
so that is the horizontal asumptote
http://www.wolframalpha.com/input/?i=(3x%5E2%2B8x-10)+%2F+(x%5E2%2B7x%2B12)+for+-10+%3C%3D+x+%3C%3D+10
= (3x^2+8x-10) / (x+3)(x+4)
no holes, since y is never 0/0
vertical asymptotes where the denominator is zero: x = -3,-4
as x gets huge, y -> 3x^2/x^2 = 3
so that is the horizontal asumptote
http://www.wolframalpha.com/input/?i=(3x%5E2%2B8x-10)+%2F+(x%5E2%2B7x%2B12)+for+-10+%3C%3D+x+%3C%3D+10
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