Find a cubic function y = ax3 + bx2 + cx + d whose graph has horizontal tangents at the points (−2, 8) and (2, 2)

Answer 1

Hi Stephanie,

Since there are 4 unknowns the idea is to find 4 equations involving a,b,c,d and use techniques for solving systems of equations to determine the value of those variables.

You will need to find the derivative y' in terms of a,b,c (y'=3a$x^2$ + 2bx +c) and also use the original function. When x=2, y = 2, y'=0. When x=-2, y=8, y'=0.

For example when x=2, y = 2 results in: 8a + 4b + 2c + d = 2.

When x = 2, y' = 0 results in: 12a + 4b + c = 0.

This graph shows the answer:

https://www.desmos.com/calculator/iilvbtnmtz

You might want to try working towards the solution before looking at the graph.

Answer 2

y' = 3ax^2 + 2bx + c

-8 a + 4b - 2c + d = 8

8 a + 4b + 2c + d = 2

0 = 12a - 4b + c

0 = 12a + 4b + c

From the last two equations, b=0

Since,

8a + 2c + d = 2

-8a -2c +d = 8

then d = 5

With these findings, the system becomes:

-8a - 2c = 3

12a + c = 0 ---> c = -12a

-8a - 2(-12a) = 3

-8a + 24a = 3

16a = 3

a = 3/16

c = -12*3/16 = -36/16 = -18/8 = -9/4

The cubic function is

y = (3/16)x^3 - (9/4)x + 5

with derivative

y' = (9/16)x^2 - 9/4

Maximizing: 0 = (9/16)x^2 - 9/4

0 = x^2 - 4 <--- multiplies both sides by 16/9

X = +- 2

AND

f(2) = (3/16)*8 - (9/4)(2) + 5 = 3/2 - 9/2 + 5 = -3 + 5 = 2

f(-2) = (3/16)* -8 - (9/4)(-2) + 5 = -3/2 + 9/2 + 5 = 3+5 = 8