Write an equation for a rational function with: Vertical asymptotes at x = 2 and x = -2 x intercepts at x = 1 and x = -4 y intercept at 4 y=

Question

Write an equation for a rational function with:

1. Vertical asymptotes at x = 2 and x = -2

1. x intercepts at x = 1 and x = -4

1. y intercept at 4

y= ?

Answer 1

1.      To get vertical asymptotes at x=2 and x=-2, the DENOMINATOR of the rational function has to be (x-2)(x+2). The reason is that when x=2 or x=-2, we would get division by zero, which is undefined, and causes the vertical asymptote.

2.      To get x intercepts at x=1 and x=-4, the NUMERATOR of the rational function has to be (x-1)(x+4). The reason is that when x=1 or x=-4, it causes the numerator to be 0. This makes y=0, and we have an x intercept.

So far we have:

a(x-1)(x+4)

y = ---------------

(x-2)(x+2)

3. The y intercept at 4 gives us the point (0,4) which we can plug in to find a.

a(0-1)(0+4)

4 = ---------------

(0-2)(0+2)

We will get a=4, so the final answer is:

4(x-1)(x+4)

y = ---------------

(x-2)(x+2)

Answer 2

Z C,

This is all about understanding fractions--not easy.

Make sure you understand why a fraction is undefined if the denominator is zero

Make sure you understand why a fraction equals zero when the numerator is zero (and the denominator is not zero)

Make sure you understand vertical asymptotes and x&y intercepts.

Here is an example: if the numerator is 10*(x-5)(x+2), and the denominator is (x-1)(x+1) then you should see vertical asymptotes when x=1 and when x=-1 because these give division by zero, and we can't factor these terms out to get a "hole" instead of a vertical asymptote

the numerator gives us x-intercepts when x= 5 and x=-2 because the fraction is zero and for the y-intercept, we let x=0 and get 100

hope this helps--Andrew F.