Question

I NEED HHHHHEEEEEELLLLLLPPPPP!!!!!!!!

Answer 1

With what?

Answer 2

The function f is a ratio of quadratic functions and has a vertical asymptote of x=4 and one x-intercept, x=1. It is known that f has a removable discontinuity at x=-1 and \[\lim_{x \rightarrow -1}f(x)=-2\]. Evaluate
(a) f(0)
(b)\[\lim_{x \rightarrow \infty}f(x)\]

Answer 3

@jim_thompson5910

Answer 4

@Directrix

Answer 5

@mathslover @mathstudent55

Answer 6

@psymon

Answer 7

Okay, so vertical asymptote at x = 4 means we have a factor of (x-4) in the bottom. A removable discontinuity at x = -1 means we have a factor of (x+1) in the top and the bottom. Because a removable discontinuity is when an undefined factor cancels. A lone x intercept of x = 1 means we have a factor of (x-1) in the top. So with that given info, we so far have:

\[\frac{ (x-1)(x+1) }{ (x-4)(x+1) }\] Now the next condition is f(-1) = -2 But if we plug in -1, we have the (x+1) factors cancel and we only get (-2/-5). Now in order for this to become -1 as an end result, the numerator must become 10 somehow. The only way this happens is if that -2 is mltiplied by -5. So this means Ill just tack on a -5 multiple on top to get:

\[\frac{ -5(x+1)(x-1) }{ (x-4)(x+1) }\] So this is the function we want that satisfies all the conditions. Then from here, lim as x -> 0 means f(0), which is
\[\frac{ -5(1)(-1) }{ (-4)(1) }= -\frac{ 5 }{ 4 } \]As for limit going to infinity, we check the degree of the numerator and the denominator. If the degree of the numerator and the denominator match, then the limit to infinity is just the ratio of the coefficients of the highest power term. Now if we were to multiply this out, x^2 would be the highest power on bottom with a coeffiicient of 1, and the highest power on top would be x^2, but with a coefficient of -5. So in the end, limi as x goes to infinity would be -5/1 = -5, the ratio of the coefficients of the highest powers top and bottom : )

Answer 8

^I knew he'd come to the rescue