Question

find all vertical asymptotes of the following function.
$f(x) = \frac{3x^2 - 6x - 144}{2x^2 - 5x}$
answer attempt 3 out of 5
one vertical asymptote
$\frac{3}{2}$

Explanation:

Step1: Factor numerator and denominator

Factor numerator: \(3x^2 - 6x - 144 = 3(x^2 - 2x - 48)=3(x - 8)(x + 6)\)
Factor denominator: \(2x^2 - 5x = x(2x - 5)\)

Step2: Simplify the function

Cancel common factors (none here, since numerator factors are \(x - 8, x + 6\) and denominator factors are \(x, 2x - 5\)).

Step3: Find values that make denominator zero

Set denominator \(x(2x - 5)=0\).
Solve \(x = 0\) or \(2x - 5 = 0 \implies x=\frac{5}{2}\).

Step4: Check if these make numerator zero

• For \(x = 0\): Numerator \(3(0 - 8)(0 + 6)= -144

eq0\).

• For \(x=\frac{5}{2}\): Numerator \(3(\frac{5}{2}-8)(\frac{5}{2}+6)=3(-\frac{11}{2})(\frac{17}{2})

eq0\).

So vertical asymptotes are at \(x = 0\) and \(x=\frac{5}{2}\). Wait, the original attempt was wrong. Let's recheck.

Wait, the function is \(f(x)=\frac{3x^2 - 6x - 144}{2x^2 - 5x}\). Factor numerator: \(3(x^2 - 2x - 48)=3(x - 8)(x + 6)\). Denominator: \(x(2x - 5)\). No common factors. So vertical asymptotes occur where denominator is zero, i.e., \(x = 0\) or \(2x - 5 = 0 \implies x=\frac{5}{2}\). So there are two vertical asymptotes, not one. The initial attempt was incorrect.

Answer:

The vertical asymptotes are \(x = 0\) and \(x=\frac{5}{2}\) (Note: The original attempt's answer was wrong; the correct vertical asymptotes are at \(x = 0\) and \(x=\frac{5}{2}\)).