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Find the equations of the horizontal asymptotes and the vertical asymptotes of f(x): f(x)= (7x^2+7x+10)/(2x^+11x-63) thanks
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on the denominator do you mean 2x^2
yes, sorry about that.
okay the vertical asymptotes are the lines x=a where a is a zero of the denominator. In this case the denominator factors to (x+9)(2x-7), therefore the vert. asymptotes occur at -9 and 2/7
horizontal asy is y=7/2
Now, as for the horizontal, it can be determined by the leading terms of both the numerator and denominator. More, specifically, i this case the exponents are the same (7x^2 and 2x^2), therefore we take only the ratio of the coefficients, thus the hor. asymptote is a y=7/2, as ictrees so eloquently put it
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HA = 7/2 VA = -9 and 7/2. thanks guys.
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