Precalculus Exam 2

Question

Write an equation for a rational function with:

Vertical Asymptotes at x=2 and x=4

x intercepts at x=5 and x=-1

y intercept at 6

y=

Answer 1

The vertical asymptotes occur where the denominator equals 0 because the function is undefined when the denominator is zero so have the terms (x-2) and (x-4) in the denominator

At the x intercepts the function equals zero so (x-5) and (x+1) are in the numerator

So we have the equation y = a(x-5)(x+1)/(x-2)(x-4)

Use the point of the y intercept (0,6) to solve for a

6=a(-5)(1)/(-2)(-4)

6= a(-5)/8

a=-48/5 so equation is y=(-48/5)(x-5)(x+1)/(x-2)(x-4)

Answer 2

y = a(x - 5)(x + 1)/(x - 2)(x - 4). To get a we use y-ntercept (0,6), so 6 = a(0 - 5)(0 + 1)/(0 - 2)(0 - 4); - 5a/8 = 6;

a = - 48/5. Now y = - 48(x² - 4x - 5)/[5(x² - 6x - 8)]