To find the polynomial, we essentially just need to find values for a, b, c, and d. we would need 4 equations. Since (-2,6) is a point on f(x), we know f(-2)=6. We can rewrite that using our formula for f(x), and plugging in -2 for x:
ax^3 + bx^2 + cx + d = 6, x = -2
-8a + 4b - 2c + d = 6
Similarly, since (2,6) is a point, we know f(2)=6, so we have:
ax^3 + bx^2 + cx + d = 3, x = 2
8a + 4b + 2c + d = 6
Now, since there are horizontal tangents at those two points, we know the derivative is zero at those points. So let's first just find the derivative of f(x):
f(x) = ax^3 + bx^2 + cx + d
f`(x) = 3ax^2 + 2bx + c
The first point has an x-coordinate of -2, and the second has an x-coordinate of 2, so we have:
f`(-2) = 0
3ax^2 + 2bx + c = 0, x = -2
12a - 4b + c = 0
f(2) = 0
3ax^2 + 2bx + c = 0, x = 2
12a + 4b + c = 0
Now we have the following 4 equations:
-8a + 4b - 2c + d = 6
8a + 4b + 2c + d = 0
12a - 4b + c = 0
12a + 4b + c = 0
So the solution is:
a = 3/16
b = 0
c = -9/4
d = 3
Plugging in those values to f(x) = ax^3 + bx^2 + cx + d, we get a final answer of:
y = (3/16)x^3 - (9/4)x + 3