# How do you find the end behavior and state the possible number of x intercepts and the value of the y intercept given h(x)=-7x^4+2x^3-3x^2+6x+4?

##### 1 Answer
Dec 16, 2016

x-intercepts ( y = 0 ): near $- {0.5}_{-}$ and ${1}_{+}$. y-intercept ( x = 0 ): 4.
As $x \to \pm \infty , y \to - \infty$.

#### Explanation:

y = h(x)

$y \left(9\right) y \left(1\right) \mathmr{and} y \left(0\right) y \left(- \frac{1}{2}\right)$ are less than 0.

So, zeros are bracketed in

( 1, 2) and (-0.5m 0)

$y ' = - 28 {x}^{3} + 6 {x}^{2} - 6 x + 6$.

$y ' \left(0\right) y ' \left(1\right) < 0$.

So a zero of y'is bracketed in (0, 1).

$y ' ' = - 84 {x}^{2} + 12 x - 6 = - 6 \left(14 {\left(x - \frac{3}{14}\right)}^{2} + \frac{5}{14}\right) < 0$

So, yi is a decreasing function.

It is now evident that there are only 2 x-intercepts.

The graph illustrates all these aspects.

x-intercepts ( zeros of y ) near $- {0.5}_{-}$ and ${1}_{+}$.

y-intercept ( x = 0 ): 4.

As $x \to \pm \infty , y \to - \infty$.

x-intercepts ) y = 0 ) near $- {0.5}_{-}$ and ${1}_{+}$. y-intercept ( x = 0 ): 4.

As $x \to \pm \infty , y \to - \infty$.

graph{-7x^4+2x^3-3x^2+6x+4 [-2, 2, -11.54, 11.45]}