*Originally posted by UserChevy*

**I have this math problem that I am stuck on. You have to find f(x) if the roots are x=(1/7), -6+3i, -6-3i, ((-5+sqrt(7))/(2)), and ((-5-sqrt(7))/(2)), and the y-intercept is (0,-162) and f(-5) does not exist. You also have to find the oblique asymptote and explain why f(-5) does not not exist. I think that I found the correct function: f(x)=5.6x^5-40 ...[text shortened]... find the oblique asymptote or why f(-5) does not exist? Any help would be greatly appreciated.**

i think the equation you are looking for can be found from a few steps:

1. let the product of all the roots = g(x) (ex. (7x-1)(x^2+6x+15)(4x^2+20x+18) = g(x) )

in other words, let each of the roots = zero, rearrange to put all the terms on one side (i.e. x=1/7 becomes 7x-1=0) and then using the reverse of the zero-product rule, you can find a polynomial that gives you zero.

now, for this equation, g(0) = -270 which is not -162. so you have to multiply through by a constant multiplier, in this case let f(x) = 162/270 * g(x). now we have the first two parts in order.

for f(-5) not to exist, the denominator must -> 0 as x->-5, in other words, put an (x+5) in the denominator.

and as far as an oblique asymptote goes, there are a number of possible answers. if there is only an (x+5)^1 power in the denominator, then there's no asymptote so the function rises to infinity. if there is a higher degree (for example (x+5)^15 ) in the denominator, then as x moves toward positive infinity, a horizontal asymptote of y=0 appears.

however, if the degree in the numerator is one greater than the degree of the denominator (our roots give us a 5th degree polynomial beginning with 28x^5) an oblique asymptote will be apparent as x approaches infinity. so... let the denominator be (x+5)^4, and do a little polynomial division. you will get a linear term and a fraction that goes to zero. so the oblique asymptote will be y = (linear term of the division)

i hope this helps i'm just too lazy to do the actual work for the last part and hopefully you'll know follow what i'm trying to say