I will give 50 points
The function f(x) is a cubic function and a limited table of values is provided below. Write the equation of the cubic polynomial in standard form.

f(x) = x³+2x²-16x-32

Step-by-step explanation:

f(x) = 0 when x = -4, -2, 4

So these are the roots

Factors are (x+4)(x+2)(x-4)

f(x) = [(x+4)(x-4)](x+2)

f(x) = (x²-16)(x+2)

f(x) = x³+2x²-16x-32

f(x) = 1x^3 + 2x^2  -16x - 32

Step-by-step explanation:

Looking at the first row of the table, I see f(x) = -27 when x = -5.  We know that -27 is the cube of -3.  So, try adding 2 to x = -5, obtaining -3.  The result (-5 + 2)^3 equals -27, as we wanted.

Unfortunately the proposed function f(x) = (x + 2)^3 does not "work" for other x values.

The governing polynomial involves four coefficients:  f(x) = ax^3 + bx^2 + cx + d.

We can pick any four x values and the corresponding y-values and come up with four versions of the polynomial f(x) = ax^3 + bx^2 + cx + d:

(1)  Choose x = -4 and f(x) = 0 from the second line in the table.  Then

     0 = a(-4)^3 + b(-4)^2 + c(-4) + d, or -64a + 16b - 4c + d = 0, and

(2) Choose x = 0 and y  = -32.  Then -32 = f(x) = a(0)^3 + b(0)^2 + c(0) + d.  Then we have -32 = 0 + 0 + 0 + d, or d = -32.

(3) Choose x = 4 and y = 0.  Then we have 0 = 64a + 16b + 4c - 32.

(4) Choose x = 5 and y = 63.  Then 63 = 125a + 25b + 5c - 32

Now we have four equations in four unknowns {a, b, c, d}.

Let's simplify these four equations, as follows:

(2) We have already seen that d = -32.  Therefore,

(3) becomes 64a + 16b + 4c = 32

(4) becomes 125a + 25b + 5c = 95.  We need one more equation.

(5) will be based on (2, -48):  -48 = 2^3a + 2^2b + 2c - 32, or

                                                 -16 = 8a + 4b + 2c

Now we have a system of linear equations to solve:

  8a + 4b + 2c = -16

125a +25b+5c = 95

 64a  + 16b + 4c = 32

Here I used the MATRIX functions of my old TI-83Plus calculator to find the values of {a, b, c}.  They are:

a=1, b=2 and c= -16

and so the general 3rd order polynomial becomes

f(x) = 1x^3 + 2x^2  -16x - 32

walter needs to make at least 5 more cookies but no more than 40. 5 ≥ x ≥ 40;

step-by-step explanation:

walter has to make a minimum of 20 cookies. he already has 15 cookies made, so he must make 5 more to meet the minimum.

the maximum amount of cookies he can make is 55. he already has 15, so he has to make no more than 40 more cookies to meet the maximum.