Question

score: 5/18 answered: 5/18 question 6 consider the rational function $g(x)=\frac{4x + 36}{x^{2}+0x - 81}$ factor the numerator to get factor the denominator to get the rational function $f$ has a vertical asymptote at $x=$ the rational function $f$ has a hole when $x=$ question help: message instructor

Explanation:

Step1: Factor the numerator

Factor out the greatest - common factor from \(4x + 36\). The GCF of 4 and 36 is 4. So, \(4x+36 = 4(x + 9)\).

Step2: Factor the denominator

We have the quadratic expression \(x^{2}-81\). Using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a=x\) and \(b = 9\), we get \(x^{2}-81=(x + 9)(x - 9)\).

Step3: Find the vertical asymptote

A vertical asymptote occurs where the denominator of a rational function is zero and the numerator is non - zero. Set the factored denominator \((x + 9)(x - 9)=0\). Solving \(x+9 = 0\) gives \(x=-9\) and solving \(x - 9=0\) gives \(x = 9\). But when \(x=-9\), the numerator is also zero. After canceling out the common factor \((x + 9)\) from the numerator and denominator, the non - canceled factor in the denominator \(x - 9=0\) gives the vertical asymptote. So the vertical asymptote is \(x = 9\).

Step4: Find the hole

A hole occurs when a factor in the numerator and a factor in the denominator cancel out. The common factor of the numerator and denominator is \(x + 9\). Setting \(x+9 = 0\), we get \(x=-9\). So there is a hole at \(x=-9\).

Answer:

Factor the numerator to get \(4(x + 9)\)
Factor the denominator to get \((x + 9)(x - 9)\)
The rational function \(f\) has a vertical asymptote at \(x = 9\)
The rational function \(f\) has a hole when \(x=-9\)