Finding Vertical Asymptotes for Rational Functions

Q1. Find the equation(s) of vertical asymptote(s) for the function $f(x) = \frac{x-1}{x^2+9x-10}$, if there are any.

Background

Topic: Rational Functions & Asymptotes

This question tests your understanding of how to find vertical asymptotes for rational functions. Vertical asymptotes occur where the denominator of the function equals zero (and the numerator does not also equal zero at that point).

Key Terms and Formulas

• Vertical Asymptote: A line $x = a$ where the function approaches infinity or negative infinity as $x$ approaches $a$.

• Rational Function: A function of the form $f(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials.

• To find vertical asymptotes: Set the denominator $Q(x)$ equal to zero and solve for $x$.

Step-by-Step Guidance

1. Write the denominator of the function: $x^2 + 9x - 10$.

2. Factor the denominator: $x^2 + 9x - 10 = (x + 10)(x - 1)$.

3. Set each factor equal to zero to find potential vertical asymptotes: $x + 10 = 0$ and $x - 1 = 0$.

4. Check if the numerator $x - 1$ is also zero at these points. If both numerator and denominator are zero at the same $x$ value, it is not a vertical asymptote (it may be a hole).

Try solving on your own before revealing the answer!

Final Answer: $x = -10$

At $x = -10$, the denominator is zero and the numerator is not zero, so there is a vertical asymptote at $x = -10$. At $x = 1$, both numerator and denominator are zero, so this is not a vertical asymptote (it is a removable discontinuity).