Question

Find a cubic function f(x) = ax^3+bx^2+cx+d whose graph has horizontal tangents at the points (-2, 6) and (2, 0)

Answer 1

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