SOLUTION: Write a rational function f(x) with asymptotes at y=7,x=-4 and a hole at x=2.
So far i got f(x)= (7x^2-14)/)(x+2)(x-2))
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Question 841459: Write a rational function f(x) with asymptotes at y=7,x=-4 and a hole at x=2.
So far i got f(x)= (7x^2-14)/)(x+2)(x-2))
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
You should check x = -4 and x = 2. If x = -4, then f(-4) = 98/12 which is not undefined, so there is no asymptote there. At x = 2, f(2) = 14/0, which means there is an asymptote, not a hole. Basically, check Wolfram Alpha or some graphing utility first.
You are right that the ratio of the leading coefficients is 7 and that the numerator/denominator have the same degree. For x = -4, we want to obtain f(-4) = k/0 where k is non-zero. For x = 2, we want f(2) = 0/0 but )
to exist.
Start by putting x+4 and x-2 in the denominator. Let  = \frac{7x^2 + P(x)}{(x+4)(x-2)})
where )
has degree at most 1. Then  = \frac{28 + P(2)}{0})
and  = \frac{112 + P(-4)}{0})
. Try  = -14x)
and you will see both conditions are satisfied. Hence, an example function that satisfies all of the constraints is
 = \frac{7x^2 - 14x}{(x+4)(x-2)})
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