To find the vertical and horizontal asymptotes of the function [tex]\( s(x) = \frac{6x^2 + 1}{2x^2 + 3x - 2} \)[/tex], let's go through the process step-by-step.
### Vertical Asymptotes:
Vertical asymptotes occur where the denominator of the function is zero, as long as the numerator is not zero at those points as well.
1. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
2x^2 + 3x - 2 = 0.
\][/tex]
2. Solve the quadratic equation:
Factoring the quadratic equation [tex]\( 2x^2 + 3x - 2 = 0 \)[/tex]:
[tex]\[
2x^2 + 4x - x - 2 = 0 \implies (2x-1)(x+2) = 0.
\][/tex]
3. Set each factor to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
2x - 1 = 0 \implies x = \frac{1}{2},
\][/tex]
[tex]\[
x + 2 = 0 \implies x = -2.
\][/tex]
Therefore, the vertical asymptotes are:
[tex]\[ x = -2 \text{ and } x = \frac{1}{2}. \][/tex]
### Horizontal Asymptote:
Horizontal asymptotes are determined by the behavior of [tex]\( s(x) \)[/tex] as [tex]\( x \)[/tex] approaches infinity or negative infinity.
1. Compare the degrees of the polynomial in the numerator and denominator:
- The degree of the numerator [tex]\( 6x^2 + 1 \)[/tex] is 2.
- The degree of the denominator [tex]\( 2x^2 + 3x - 2 \)[/tex] is 2.
2. When the degree of the numerator and the degree of the denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. Here, the leading coefficients are 6 (numerator) and 2 (denominator):
[tex]\[
\text{Horizontal Asymptote} = \frac{6}{2} = 3.
\][/tex]
Therefore, the horizontal asymptote is:
[tex]\[ y = 3. \][/tex]
### Final Answer:
- Vertical Asymptotes: [tex]\( x = -2, \frac{1}{2} \)[/tex]
- Horizontal Asymptote: [tex]\( y = 3 \)[/tex]
