Question

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`?

Answer 1

x-intercepts ( y = 0 ): near `-0.5_-` and `1_+`. y-intercept ( x = 0 ): 4.
As `x to +-oo, y to -oo`.

Explanation:

y = h(x)

`y(9)y(1) and y(0)y(-1/2)` are less than 0.

So, zeros are bracketed in

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

`y'=-28x^3+6x^2-6x+6`.

`y'(0)y'(1)<0`.

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

`y''=-84x^2+12x-6=-6(14(x-3/14)^2+5/14)<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 +-oo, y to -oo`.

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

As `x to +-oo, y to -oo`.

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