question determine if the function below has a horizontal or a slant asymptote, and explain why. f(x)=(-5 +…

Explanation:

Step1: Identify degrees of numerator and denominator

The degree of the numerator $-5 + x^{2}-11x - 6x^{3}$ is 3 (highest - power of $x$ is 3), and the degree of the denominator $-x + 4+2x^{2}$ is 2 (highest - power of $x$ is 2).

Step2: Determine the type of asymptote

Since the degree of the numerator (3) is greater than the degree of the denominator (2), the function has a slant asymptote. We find the slant asymptote by long - dividing the numerator by the denominator. Dividing $-6x^{3}+x^{2}-11x - 5$ by $2x^{2}-x + 4$: [ \begin{align*} -6x^{3}+x^{2}-11x - 5&=-3x(2x^{2}-x + 4)+(-3x)(-x)+(-3x)(4)+x^{2}-11x - 5\ &=-3x(2x^{2}-x + 4)+3x^{2}-12x+x^{2}-11x - 5\ &=-3x(2x^{2}-x + 4)+4x^{2}-23x - 5\ &=-3x(2x^{2}-x + 4)+2(2x^{2}-x + 4)+2(-x)+2(4)-23x - 5\ &=(-3x + 2)(2x^{2}-x + 4)-2x+8-23x - 5\ &=(-3x + 2)(2x^{2}-x + 4)-25x + 3 \end{align*} ] So, $f(x)=\frac{-6x^{3}+x^{2}-11x - 5}{2x^{2}-x + 4}=-3x + 2+\frac{-25x + 3}{2x^{2}-x + 4}$. As $x$ approaches infinity (positive or negative), the value of $\frac{-25x + 3}{2x^{2}-x + 4}$ approaches 0. So, as $x$ approaches infinity (positive or negative), the value of $y$ approaches $-3x + 2$.

Answer:

The function has a slant asymptote because the degree of the numerator is greater than the degree of the denominator. As $x$ approaches infinity (positive or negative), the value of $y$ approaches $-3x + 2$.