To find the horizontal and vertical asymptotes for the given rational function:

$y = \frac{2x^2 + x - 1}{x^2 + x - 90}$

1. Horizontal Asymptote

Horizontal asymptotes depend on the degrees of the polynomials in the numerator and the denominator.

• The degree of the numerator $2x^2 + x - 1$ is 2.
• The degree of the denominator $x^2 + x - 90$ is also 2.

When the degrees of the numerator and denominator are the same, the horizontal asymptote is found by taking the ratio of the leading coefficients.

• Leading coefficient of the numerator: $2$
• Leading coefficient of the denominator: $1$

Thus, the horizontal asymptote is:

$y = \frac{2}{1} = 2$

2. Vertical Asymptote

Vertical asymptotes occur where the denominator equals zero (and the numerator does not simultaneously equal zero at the same value).

Set the denominator equal to zero and solve for $x$:

$x^2 + x - 90 = 0$

This quadratic equation can be solved using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = 1$, and $c = -90$. Substituting these values in:

$x = \frac{-1 \pm \sqrt{1 + 360}}{2} = \frac{-1 \pm \sqrt{361}}{2} = \frac{-1 \pm 19}{2}$

So, the solutions are:

$x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{-20}{2} = -10$

Therefore, the vertical asymptotes are:

$x = 9 \quad \text{and} \quad x = -10$

Summary

• Horizontal Asymptote: $y = 2$
• Vertical Asymptotes: $x = 9$ and $x = -10$

Would you like more details or have any questions?

Related Questions:

1. How do you find oblique asymptotes?
2. What happens if the degree of the numerator is higher than the denominator?
3. Can there be a situation where a function has no asymptotes at all?
4. How can horizontal asymptotes be used to understand the end behavior of a function?
5. What is the significance of the roots of the numerator in determining the asymptotes?

Tip: Always check the numerator when identifying vertical asymptotes, as common factors between numerator and denominator can remove potential asymptotes.