Question

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reattempt last question below, or select another question
suppose ( f(x)=\frac{x + 8}{x^{2}-4x} )
a. what value(s) of ( x ) make the denominator of ( fx^{2}-4x ) have a value of 0? if there is more than one value, enter your answer as a comma - separated list
( x=) preview ) = undefined (unmatched parens)
b. using a graphing calculator, graph ( f ) and verify that a vertical asymptote occurs at the value(s) of ( x ) that you listed in part (a)
points possible: 2
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Explanation:

Step1: Set the denominator equal to zero

We have \(f(x)=\frac{x + 8}{x^{2}-4x}\). For the denominator \(x^{2}-4x\) of \(f(x)\) to be zero, we set \(x^{2}-4x = 0\).

Step2: Factor the equation

Factor out an \(x\) from \(x^{2}-4x\). So, \(x(x - 4)=0\).

Step3: Solve for \(x\)

Using the zero - product property \(ab = 0\) implies \(a=0\) or \(b = 0\). Here, if \(x=0\) or \(x - 4=0\) (i.e., \(x = 4\)), the equation \(x(x - 4)=0\) is satisfied.

Answer:

\(0,4\)