Intercepts of Rational Functions

Problem 112: Given the function

(1) $\begin{equation*} f(x) = \frac{x^2-3x-40}{x^2-9x+20},\end{equation*}$

find

(a) the y-intercept, and

(b) x- intercept.

Solution: First, we need to know the definition of the y-intercept and x-intercept. That is, the y-intercept is given by the point $(x,y) = (0,y)$ and the x-intercept is given by the point $(x,y) = (x,0)$ .

(a) The y-intercept is then

(2) $\begin{align*} y & = \frac{x^2-3x-40}{x^2-9x+20} \\ y & = \frac{(0)^2-3(0)-40}{(0)^2-9(0)+20} \\ & = \frac{-40}{20} \\ & = -2.\end{align*}$

That is, the y-intercept is given by $(x,y) = (-2,0)$ .

(b) The x-intercept is given by

(3) $\begin{align*} f(x) & = \frac{x^2-3x-40}{x^2-9x+20} \\ 0 & = \frac{x^2-3x-40}{x^2-9x+20}. \end{align*}$

Note that in order for the rational function to be equal to zero, we will need ONLY the numerator to be equal to zero because if the denominator is equal to zero, the rational function is undefined. That is,

(4) $\begin{align*} 0 & = x^2 - 3x- 40 \\ & = (x-8)(x+5) \\ & \Rightarrow x = 8 \ \text{or} \ x = -5.\end{align*}$

Hence, the x-intercepts is given by $(x,y) = (0,8) \ \text{and} \ (x,y) = (0,-5)$ .

Leave a Reply Cancel reply