Question

find all vertical asymptotes of the following function.
$f(x) = \frac{x^2 - 8x + 15}{3x^2 - 9x}$

Explanation:

Step1: Factor numerator and denominator

Factor the numerator \(x^{2}-8x + 15\): we need two numbers that multiply to \(15\) and add to \(-8\), which are \(-3\) and \(-5\). So \(x^{2}-8x + 15=(x - 3)(x - 5)\).

Factor the denominator \(3x^{2}-9x\): factor out \(3x\), we get \(3x^{2}-9x = 3x(x - 3)\).

So the function becomes \(f(x)=\frac{(x - 3)(x - 5)}{3x(x - 3)}\).

Step2: Simplify the function (remove common factors)

We can cancel out the common factor \((x - 3)\) for \(x
eq3\) (since we can't divide by zero). So the simplified function is \(f(x)=\frac{x - 5}{3x}\) for \(x
eq3\).

Step3: Find values that make the original denominator zero (and check if they are vertical asymptotes)

The original denominator is \(3x^{2}-9x=3x(x - 3)\). Set the denominator equal to zero: \(3x(x - 3)=0\). Solving for \(x\), we get \(x = 0\) or \(x=3\).

Now, we check the limit as \(x\) approaches these values.

• For \(x = 3\): We already canceled the \((x - 3)\) factor, and the simplified function \(\frac{x - 5}{3x}\) is defined at \(x = 3\) (since plugging in \(x = 3\) gives \(\frac{3-5}{3\times3}=\frac{-2}{9}\)), so \(x = 3\) is a hole (removable discontinuity), not a vertical asymptote.

• For \(x=0\): As \(x\to0\), the numerator of the simplified function \(\frac{x - 5}{3x}\) approaches \(-5\) and the denominator approaches \(0\). So \(\lim_{x\to0}\frac{x - 5}{3x}\) will be either \(+\infty\) or \(-\infty\) (depending on the side). So \(x = 0\) is a vertical asymptote.

Answer:

The vertical asymptote is \(x = 0\)