# How do you describe the end behavior of y=8-x^3-2x^4?

May 28, 2018

See explanation.

#### Explanation:

To describe end behavior of a function you need to calculate the limits:

and

## ${\lim}_{x \to + \infty} f \left(x\right)$

Here you get:

${\lim}_{x \to \pm \infty} \left(8 - {x}^{3} - 2 {x}^{4}\right) = {\lim}_{x \to \pm \infty} \left({x}^{4} \cdot \left(- 2 - \frac{1}{x} + \frac{8}{x} ^ 3\right)\right)$

If $x$ goes to $\pm \infty$ the fractions in the brackets go to zero, and ${x}^{4}$ goes to $+ \infty$, so the whole expression goes to $- \infty$. Therfore we can say that as $x$ goes to $- \infty$ or $+ \infty$ the function goes to $- \infty$

May 28, 2018

as $x \to \infty , f \left(x\right) \to - \infty$
as $x \to - \infty , f \left(x\right) \to - \infty$

#### Explanation:

$y = 8 - {x}^{3} - 2 {x}^{4}$

To determine the "end behavior" we need only consider the highest degree term because as $x \to \infty$ the highest degree term dominates the function's behavior:

$- 2 {x}^{4}$

${x}^{4}$ has very similar behavior to ${x}^{2}$ just grows more rapidly.

since the coefficient is negative the function's end behavior is decreasing, so we have determined:

$f \left(x\right) = - 2 {x}^{4} - {x}^{3} + 8$

as $x \to \infty , f \left(x\right) \to - \infty$
as $x \to - \infty , f \left(x\right) \to - \infty$

graph{-2x^4-x^3+8 [-10.545, 9.455, -1.4, 8.6]}