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#?

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Answer 1 · 2016-12-16T03:24:36

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

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]}