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

Question

 Find a cubic function y = ax³ + bx² + cx + d whose graph has horizontal tangents at the points (-2, 6) and (2, 0)

 

I get that the function derives to:

3ax² + bx + x

 

I then need to set it to zero and sub in an x value

 

However, this never gets me the right answer (see below)

 

 

#65, 3.1

ANSWER:    y = 3/16 x³ - 9/4 x² + 3

Answer 1

Horizontal tangents means that the slope of the line tangent to the cubic function is zero.  Slope of tangent line is the same as derivative.  According to first derivative test, the given points are local extremas.

 

First, lets set the derivative of the cubic function equal to zero.

 

3ax² + 2bx + c = 0

 

 

Plug in both x-values into the derivative. These are called critical points, the location of the local extremas.

 

3a(2)² + 2b(2) + c = 0

 

12a + 4b + c = 0

 

 

3a(-2)² + 2b(-2) + c = 0

 

12a - 4b + c = 0

 

 

When we plug in the x values of the points into the cubic function, we get the y values of the local extremas.

 

a(-2)³ + b(-2)² + c(-2) + d = 6

 

-8a + 4b - 2c + d = 6           

 

 

a(2)³ + b(2)² + c(2) + d = 0

 

8a + 4b + 2c + d = 0

 

 

 

The bolded equations that we just establish become a system of equations.  We use these equations to solve for a, b, c, and d.

 

12a + 4b + c = 0                       eq1

12a - 4b + c = 0                        eq2

-8a + 4b - 2c + d = 6                 eq3

8a + 4b + 2c + d = 0                 eq4

 

 

Use the substitution and elimination methods to solve the system.