Question

Which graph represents the function f(x)=x4+4x3−12x2−32x+64?

Answer 1

$f(x)=x^4+4x^3-12x^2-32x+64$

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

$f(x)=x^4+4x^3-12x^2-32x+64$

$f(x)=(-4)^4+4(-4)^3-12(-4)^2-32(-4)+64=0$

Now plug in 2 for x and check

$f(x)=(2)^4+4(2)^3-12(2)^2-32(2)+64=0$

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

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

Answer 2

The function $f(x)=x^4+4x^3-12x^2-32x+64$

has a derivative

$f^\prime(x)=(x^3+4x^2-6x-8).$

The derivative of this function is zero when $x=-4,x=-1,x=2.$

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