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f(x)= (11x^2+3x+50)/5x^2-25x

2006-12-12 00:56:46 · 2 answers · asked by susan h 1 in Science & Mathematics Mathematics

2 answers

limf(x)=11/5 with x of infinity.Then y=11/5 is horizontal asymptotes.
5x^2-25x=0
5x(x-5)=0 then x=0 or x=5
x=0 and x=5 are the equations of the vertical asymptotes.

2006-12-12 01:04:49 · answer #1 · answered by grassu a 3 · 0 0

Since the numerator and denominator are both of degree 2, then there will be a horizontal asymptote at y = 11x^2/5x^2 = 11/5

Since the denominator equals zero when x = 0 and x= 5 then x+0 and x=5 are equations of the vertical asymptotes.

2006-12-12 09:14:34 · answer #2 · answered by ironduke8159 7 · 0 0

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