determine if the function below has a horizontal or a slant asymptote, and explain why. f(x)=(-5 + x^2 - 11x…

Explanation:

Step1: Identify degrees of numerator and denominator

The numerator $-6x^{3}+x^{2}-11x - 5$ has degree 3. The denominator $2x^{2}-x + 4$ has degree 2. Since $3>2$, there is a slant asymptote.

Step2: Perform polynomial long - division

Divide $-6x^{3}+x^{2}-11x - 5$ by $2x^{2}-x + 4$. [ \begin{align*} -6x^{3}+x^{2}-11x - 5&=-3x(2x^{2}-x + 4)+(-3x)(-x)+x^{2}-11x - 5\ &=-3x(2x^{2}-x + 4)+3x^{2}+x^{2}-11x - 5\ &=-3x(2x^{2}-x + 4)+4x^{2}-11x - 5\ &=-3x(2x^{2}-x + 4)+2(2x^{2}-x + 4)+2x-13 \end{align*} ] The quotient of the long - division is $-3x + 2$. As $x\to\pm\infty$, the remainder term $\frac{2x - 13}{2x^{2}-x + 4}\to0$, and $y$ 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 $-3x + 2$.