Determine the vertical asymptotes of $f(x)=\frac{2x^2-3x+5}{x^2-5x+6}$

Expert intro: Vertical asymptotes of a rational function occur where the denominator is zero after any common factors are canceled. The numerator here does not factor with the denominator, so the denominator alone determines the asymptotes.

Detailed walkthrough

Step 1: Factor the denominator

$x^2-5x+6=(x-2)(x-3).$

So the function is

$f(x)=\frac{2x^2-3x+5}{(x-2)(x-3)}.$

Step 2: Check for cancellation

Vertical asymptotes can disappear if a factor cancels with the numerator. Here, the numerator is

$2x^2-3x+5,$

which does not factor as $(x-2)$ or $(x-3)$, and neither factor cancels.

Step 3: Identify where the denominator is zero

Set each denominator factor equal to zero:

$x-2=0 \Rightarrow x=2,$

$x-3=0 \Rightarrow x=3.$

Final answer

The vertical asymptotes are

$\boxed{x=2 \text{ and } x=3}.$

These are the values where the function is undefined and the graph shoots up or down without bound.

💡 Pitfall guide

Do not forget to check whether any factor cancels first. If a factor cancels, that x-value is a hole, not a vertical asymptote. Here, no cancellation occurs, so both zeros of the denominator are asymptotes.

🔄 Real-world variant

If the numerator had contained a factor $(x-2)$, then $x=2$ would be a removable discontinuity instead of a vertical asymptote. If the denominator had repeated factors, such as $(x-2)^2$, the vertical asymptote would still be at $x=2$, but the graph’s behavior near it could be different.

🔍 Related terms

rational function, vertical asymptote, removable discontinuity