# Precalc

Find an equation of a rational function that satisfies the given conditions.
Vertical asymptote: x = 4
Horizontal asymptote: y = 0
X-intercepts: none
Y-intercepts (0, -1/4)
Multiplicity of 1
Removable discontinuity: x = 1
End behavior: x --> infinity, f(x) --> -infinity; x --> -infinity, f(x) --> infinity.

I can graph this, but I am unsure how to turn this into a function. Please help? Thanks

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1. You want f(±∞) = +∞
This is impossible, since there is a horizontal asymptote at y=0.

hole and vertical asymptote:

(x-1) / (x-4)(x-1)

This also gives y=0 as the vertical asymptote, since the denominator has higher degree than the numerator.

not sure what has the multiplicity of 1. That is usually reserved for the roots, but you say there are no x-intercepts.

Now f(0) = (-1)/(-4)(-1) = -1/4 as desired.

So, apart from the end behavior, this looks good. See

http://www.wolframalpha.com/input/?i=(x-1)+%2F+((x-4)(x-1))

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2. Okay, thank you, Steve

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