Question

What are the zero(s) of the function f(x)=$\frac{4x^{2}-36x }{x-9}$?
a.) x = -9
b.)x = 0
c.)x = 9
d.)x = 0 and x = 9

Answer 1

Answer:

D

Step-by-step explanation:

The function will be zero when the numerator is equal to 0. So, set the numerator equal to 0 and solve:

$0 = 4 {x}^{2} - 36x$

I will use the quadratic formula:

$\frac{ 36 + \sqrt{ {36}^{2} - 4(4)(0)} }{2(4)} \\ \frac{ 36 + 36}{8} = 9$

Therefore, x = 9 is a 0. Let's check with subtracted root now:

$\frac{ 36 - \sqrt{ {36}^{2} - 4(4)(0)} }{2(4)} \\ \frac{ 36 - 36}{8} = 0$

It appears 0 is also a root.

Therefore, the answer is D.