# How do you sketch the general shape of f(x)=-x^4+3x^2-2-5x using end behavior?

Dec 2, 2016

As $x \to \pm \infty , y \to - \infty$. The zenith (y' = 0 ) is close to (1.5, 7.8125).
There are two points of inflexion at $x = \pm \sqrt{2}$. See the illustrative graph.

#### Explanation:

f(x) has alternate signs at x = 0, -1 and -2. So, the graph cuts xaxis

twice in (-2, -1).

f'=-4x^3+6x-5 and changes sign nera x = -1.5, for a turning point.

$f ' ' = - 12 {x}^{2} + 6 = 0$, for $x = \pm \sqrt{2} \mathmr{and} f ' ' ' = - 12$.

So, the points of inflexion are $\pm \left(\sqrt{2} , - 5 \sqrt{2}\right)$.

$y = f = - {x}^{4} \left(1 - \frac{3}{x} ^ 2 + \frac{5}{x} ^ 3 + \frac{2}{x} ^ 4\right) \to - \infty$, as $x \to \pm \infty$.

graph{-x^4+3x^2-5x-2 [-20, 20, -10, 10]}