Question

question find all holes of the following function. if a hole exists, write your answer as a coordinate point in simplest form. $f(x)=\frac{x^{2}-81}{x^{2}+9x}$. answer attempt 1 out of 2 add a note no holes watch video show examples

Explanation:

Step1: Factor the denominator

The denominator \(x^{2}-81=(x + 9)(x - 9)\) by the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\) where \(a=x\) and \(b = 9\). The function is \(f(x)=\frac{x^{2}}{x^{2}-81}=\frac{x^{2}}{(x + 9)(x - 9)}\). A hole occurs when a factor in the denominator can be canceled out with a factor in the numerator. Since the numerator \(x^{2}=x\times x\) has no common factors with \((x + 9)(x - 9)\), there are no holes in the function.

Answer:

No Holes