Question

what are the x-intercept(s) of the function ( f(x)=\frac{4x^2 - 36x}{x - 9} )?
( x = -9 )
( x = 0 )
( x = 9 )
( x = 0 ) and ( x = 9 )

Explanation:

Step1: Recall x-intercept definition

To find the x - intercepts of a function \(y = f(x)\), we set \(y=0\) (i.e., \(f(x) = 0\)) and solve for \(x\), with the condition that the function is defined at that \(x\) value.

Given the function \(f(x)=\frac{4x^{2}-36x}{x - 9}\), we set \(f(x)=0\):
\[
\frac{4x^{2}-36x}{x - 9}=0
\]
A fraction is equal to zero when its numerator is zero and its denominator is not zero.

Step2: Solve the numerator for zero

Solve \(4x^{2}-36x = 0\). Factor out \(4x\) from the left - hand side:
\[
4x(x - 9)=0
\]
Using the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), we have two cases:

• Case 1: \(4x=0\), then \(x = 0\).
• Case 2: \(x - 9=0\), then \(x = 9\). But we need to check the denominator. When \(x = 9\), the denominator \(x-9=9 - 9=0\), and the function \(f(x)=\frac{4x^{2}-36x}{x - 9}\) is undefined at \(x = 9\) (since division by zero is not allowed). So we discard \(x = 9\) as a solution.

So the only solution for \(f(x)=0\) is \(x = 0\).

Answer:

\(x = 0\) (the option corresponding to \(x = 0\))