Find an equation of a rational function f that satisfies the

Find an equation of a rational function f that satisfies the given conditions. vertical asymptotes: x = 7, x = 5 horizontal asymptote: y = 0 x-intercept: 1; f(0) = 2 hole at x = 2

Solution

The vertical asymptotes (and any restrictions on the domain) come from the zeroes of the denominator.

Denominator : (x+7)(x-5)

The degree is greater in the denominator than in the numerator, the y-values will be dragged down to the x-axis, and the horizontal asymptote is therefore \"y = 0\"

x intercept x=-1 which is numerator factor --> (x+1)

Holes in the graph of a rational function are generally produced by factors that are common to both the numerator and the denominator x=2

So, F(x) = (x+1)(x-2)/(x-7)(x-5)(x-2)

Given f(0) =-2 ; So lets use a constant k, F(x) = k(x+1)(x-2)/(x-7)(x-5)(x-2)

Now f(0) =-2; -2 = k*1/(-7*-5)

k = 70

F(x) = 70(x+1)(x-2)/(x-7)(x-5)(x-2)