Question

find the equation of all vertical asymptotes of the following function.
$f(x) = \frac{x + 2}{\sqrt{x^2 - 12x + 32}}$

Explanation:

Step1: Factor the denominator's radicand

First, we factor the quadratic expression inside the square root in the denominator. The quadratic is \(x^{2}-12x + 32\). We need two numbers that multiply to \(32\) and add up to \(- 12\). Those numbers are \(-4\) and \(-8\). So, \(x^{2}-12x + 32=(x - 4)(x - 8)\). So the function becomes \(f(x)=\frac{x + 2}{\sqrt{(x - 4)(x - 8)}}\).

Step2: Determine the domain restrictions (where the denominator is zero or undefined)

The denominator of a function (especially with a square root) is undefined when the expression inside the square root is negative, and also when the square root is zero (since division by zero is undefined). So we need to find where \((x - 4)(x - 8)=0\) or \((x - 4)(x - 8)<0\).

First, find the roots of \((x - 4)(x - 8)=0\). Setting each factor equal to zero gives \(x=4\) and \(x = 8\).

Now, we analyze the sign of \((x - 4)(x - 8)\) in different intervals:

• For \(x<4\): Let's take \(x = 0\). Then \((0 - 4)(0 - 8)=(-4)\times(-8)=32>0\).
• For \(4
• For \(x>8\): Let's take \(x = 9\). Then \((9 - 4)(9 - 8)=(5)\times(1)=5>0\).

The expression inside the square root \((x - 4)(x - 8)\) must be positive (since it's under a square root in the denominator, and the denominator can't be zero). So the domain of the function is \(x<4\) or \(x>8\) (because in \(4 < x<8\), the expression inside the square root is negative, so the function is undefined there, and at \(x = 4\) and \(x = 8\), the denominator is zero).

Step3: Check for vertical asymptotes

A vertical asymptote occurs at a value \(x = a\) if the function approaches \(\pm\infty\) as \(x\) approaches \(a\) from the left or right, and \(a\) is not in the domain of the function.

• As \(x\) approaches \(4\) from the left (\(x

ightarrow4^{-}\)): Since \(x<4\), \((x - 4)\) is negative and \((x - 8)\) is negative, so \((x - 4)(x - 8)\) is positive. The denominator \(\sqrt{(x - 4)(x - 8)}\) approaches \(0\) from the positive side (because the square root of a positive number approaching \(0\) is positive). The numerator \(x + 2\) approaches \(4+2 = 6\) (a non - zero constant). So \(\lim_{x
ightarrow4^{-}}\frac{x + 2}{\sqrt{(x - 4)(x - 8)}}\): as \(x
ightarrow4^{-}\), \((x - 4)
ightarrow0^{-}\) and \((x - 8)
ightarrow - 4\), so \((x - 4)(x - 8)
ightarrow0^{+}\), \(\sqrt{(x - 4)(x - 8)}
ightarrow0^{+}\), and the numerator approaches \(6\). So the limit is \(+\infty\) (since numerator is positive and denominator approaches \(0\) from the positive side).

• As \(x\) approaches \(4\) from the right (\(x

ightarrow4^{+}\)): But \(x = 4^{+}\) is in the interval \(4

• As \(x\) approaches \(8\) from the right (\(x

ightarrow8^{+}\)): Since \(x>8\), \((x - 4)\) is positive and \((x - 8)\) is positive, so \((x - 4)(x - 8)\) is positive. The denominator \(\sqrt{(x - 4)(x - 8)}\) approaches \(0\) from the positive side. The numerator \(x + 2\) approaches \(8 + 2=10\) (a non - zero constant). So \(\lim_{x
ightarrow8^{+}}\frac{x + 2}{\sqrt{(x - 4)(x - 8)}}\): as \(x
ightarrow8^{+}\), \((x - 4)
ightarrow4\) and \((x - 8)
ightarrow0^{+}\), so \((x - 4)(x - 8)
ightarrow0^{+}\), \(\sqrt{(x - 4)(x - 8)}
ightarrow0^{+}\), and the numerator approaches \(10\). So the limit is \(+\infty\).

• As \(x\) approaches \(8\) from the left (\(x

ightarrow8^{-}\)): But \(x = 8^{-}\) is in the interval \(4

Answer:

The equations of the vertical asymptotes are \(x = 4\) and \(x = 8\)