Asked by Anonymous
Write a rational function satisfying the following criteria. vertical Asymptote: x=-1, slant asymptote: y=x+2, zero of the function: x=3
I had f(x)=x^2+3x+2/x+1,
that only works for the asymptotes and not the zero can someone please help me figure how to satisfy all three requirements
I had f(x)=x^2+3x+2/x+1,
that only works for the asymptotes and not the zero can someone please help me figure how to satisfy all three requirements
Answers
Answered by
Steve
f(x)=x^2+3x+2/x+1
= (x+1)(x+2)/(x+1)
does not have a vertical asymptote. It just has a hole at x = -1
We want something with (x-3) on top, and (x+1) in the bottom, so
f(x) = (x-3)/(x+1)
has the zero and the vertical asymptote. Note that for large x, f(x) -> 1. So, we can just multiply that by (x+2) to make the slant asymptote:
f(x) = (x-3)(x+2)/(x+1)
Now, that has two zeros.
If you play around some, you can get rid of the extra zero.
http://www.wolframalpha.com/input/?i=%28x-3%29%28x%2B2%29%2F%28x%2B1%29
= (x+1)(x+2)/(x+1)
does not have a vertical asymptote. It just has a hole at x = -1
We want something with (x-3) on top, and (x+1) in the bottom, so
f(x) = (x-3)/(x+1)
has the zero and the vertical asymptote. Note that for large x, f(x) -> 1. So, we can just multiply that by (x+2) to make the slant asymptote:
f(x) = (x-3)(x+2)/(x+1)
Now, that has two zeros.
If you play around some, you can get rid of the extra zero.
http://www.wolframalpha.com/input/?i=%28x-3%29%28x%2B2%29%2F%28x%2B1%29
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