Chapter 8.3, Problem 50PPSE

Solution

To find: The $x-$ coordinate of the hole in the graph of the rational function $y=\frac{x^2-9}{2x^2-x-15}$ .

The hole of the rational function has $x-$ coordinate $3$ .

Given:

The rational function is $y=\frac{x^2-9}{2x^2-x-15}$ .

Concept:

Holes of a rational function: The holes of a rational function can be evaluated by reducing the rational function first and look for common factors in numerator and denominator and equate the common term in numerator and denominator equal to zero to get the holes of the rational function. And, if the rational function is already in reduced form, then there are no holes in the rational function.

Calculation:

Change the rational function $y=\frac{x^2-9}{2x^2-x-15}$ to reduced form and using the concept stated above set the common factor equal to zero to obtain the holes of the rational function.

  $\begin{array}{c}y=\frac{x^2-9}{2x^2-x-15}\mathrm{....}\left(\text{Given}\right)\\ y=\frac{x^2-3^2}{2x^2-6x+5x-15}\mathrm{....}\left(\text{Factorize}\right)\\ y=\frac{\left(x-3\right)\left(x+3\right)}{2x\left(x-3\right)+5\left(x-3\right)}\\ y=\frac{\left(x-3\right)\left(x+3\right)}{\left(2x+5\right)\left(x-3\right)}\\ x-3=0\mathrm{....}\left(\begin{array}{l}\text{ Set common factor }\\ \left(x-3\right)\text{ equal to zero}\end{array}\right)\\ x=3\end{array}$

Hence, the hole of the rational function has $x-$ coordinate $3$ .

Conclusion:

The hole of the rational function has $x-$ coordinate $3$ .