Respuesta :

[tex]f(x)=x^4+4x^3-12x^2-32x+64[/tex]

The degree of f(x) is 4. Also the leading coefficient is 1 and it is positive

So  as x approaches infinity then y approaches infinity

as x approaches -infinity then y approaches infinity

The first and fourth graph goes up and it satisfies the above . so we ignore the second and third graph.

Now we check the x intercepts of the first  graph

x intercepts of first graph is -4  and 2

Plug in -4 for x in f(x) and check whether we get 0

[tex]f(x)=x^4+4x^3-12x^2-32x+64[/tex]

[tex]f(x)=(-4)^4+4(-4)^3-12(-4)^2-32(-4)+64=0[/tex]

Now plug in 2 for x and check

[tex]f(x)=(2)^4+4(2)^3-12(2)^2-32(2)+64=0[/tex]

So -4  and 2  are the x intercepts that satisfies f(x)

Hence first option is the graph of [tex]f(x)=x^4+4x^3-12x^2-32x+64[/tex]





mathjohndough

The function [tex]f(x)=x^4+4x^3-12x^2-32x+64 \\[/tex]

has a derivative

[tex]f^\prime(x)=(x^3+4x^2-6x-8).\\[/tex]

The derivative of this function is zero when [tex]x=-4,x=-1,x=2.[/tex]

The derivative function is negative on the left side of -4 and positive on the right side, it is also negative on the right hand side of -1, and positive on the right hand side of positive 2. From this information we can gather that the graph of f(x)  has a minima at x=-4, a maxima at x=-1 and a minima at x=-2.

The only graph that fits this description from the given ones is the first graph


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