Answer:
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
