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

Dec 5, 2016

y-intercept ( x = 0 ) : -4. Cuts x-axis at x = -1 and again between x = -2

and x = -1. As $x \to \pm \infty , y \to - \infty$. See illustrative graph, for shape.

#### Explanation:

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

Sign changing of a functions y and y'', in an interval, reveals the

presence of zero(s) of the function, within the interval.

$y ' ' = - 12 {x}^{2} + 6 = 9 , \to x = \pm \frac{1}{\sqrt{2}} \mathmr{and} y ' ' '$ is not 0.

So, the points of inflexion ( tangent crossing curve ) are at

$x = \pm \frac{1}{\sqrt{2}}$,.

graph{y+x^4-3x^2+2x+4=0 [-20.18, 20.18, -10.1, 10.07]}

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