Find a cubic function y = ax^3 + bx^2 + cx + d whose graph has horizontal tangents at the points (−2, 8) and (2, 2).

Answer 1

If a tangent line to a curve is horizontal at a point on the curve, then the derivative must be zero at that point.

So, for y = ax³ + bx² + cx + d, then y' = 3ax² + 2bx + c

So, when x = -2, 12a - 4b + c = 0 and when x = 2, 12a + 4b + c = 0.

Adding the equations, we get 24a + 2c = 0. So, c = -12a,

Replace c by -12a in the equation 12a - 4b + c = 0 to get b = 0..

We now have y = ax³ -12ax + d.

Since the points (-2,8) and (2,2) are on the graph, we get:

-8a + 24a +d = 8

8a -24a + d = 2

Add the equations to obtain 2d = 10. So, d = 5.

So, -8a + 24a + 5 = 8

16a = 3

a = 3/16 and c = -12a = -9/4

Therefore, y = (3/16)x³ - (9/4)x + 5