Question

write an equation for a rational function with:
vertical asymptotes at $x = 2$ and $x = -1$
$x$-intercepts at $x = 4$ and $x = 5$
$y$-intercept at 5
$y = \square$
question help: video read written example

Explanation:

Step1: Set denominator from asymptotes

For vertical asymptotes at $x=2$ and $x=-1$, the denominator has factors $(x-2)$ and $(x+1)$. So denominator: $(x-2)(x+1)$

Step2: Set numerator from x-intercepts

For x-intercepts at $x=4$ and $x=5$, the numerator has factors $(x-4)$ and $(x-5)$. Let the leading coefficient be $a$, so numerator: $a(x-4)(x-5)$
The function is now: $y=\frac{a(x-4)(x-5)}{(x-2)(x+1)}$

Step3: Solve for a using y-intercept

Y-intercept is 5, so substitute $x=0, y=5$:
$5=\frac{a(0-4)(0-5)}{(0-2)(0+1)}$
Simplify right-hand side:
$5=\frac{a(-4)(-5)}{(-2)(1)}=\frac{20a}{-2}=-10a$
Solve for $a$:
$a=\frac{5}{-10}=-\frac{1}{2}$

Step4: Substitute a into function

Substitute $a=-\frac{1}{2}$ into the function:
$y=\frac{-\frac{1}{2}(x-4)(x-5)}{(x-2)(x+1)}$
We can rewrite this to eliminate the fraction in the numerator:
$y=\frac{-(x-4)(x-5)}{2(x-2)(x+1)}$ or expand if needed:
First expand numerator: $-(x^2-9x+20)=-x^2+9x-20$
Denominator: $2(x^2-x-2)=2x^2-2x-4$
So $y=\frac{-x^2+9x-20}{2x^2-2x-4}$

Answer:

$y=\frac{-\frac{1}{2}(x-4)(x-5)}{(x-2)(x+1)}$ (or expanded form $y=\frac{-x^2+9x-20}{2x^2-2x-4}$)