Derivative of a Polynomial Function

Question

Determine equations for the tangents to the cubic function y=2x^3 -3x^2 -11x +8 at the point where y=2.

Answer 1

Hi,

      Y=2 at x=-2, 1/2, and 3 and dy/dx=6x²-6x-11 so the slopes are calculated by evaluating dy/dx at -2,1/2,and 3. this will be the "m" in (y-y1)=m(x-x₁) I'll do the first of the three equations for the tangent lines.

 

x₁=-2 y₁=2 now m=6*(-2)²-6*(-2)-11 = 25 so the equation for the first tangent line is y-2=25(x+2).

 

THe other two are done in the same manor with x₁=1/2 and 3 and y₁=2 and 2 try it yourself. Let me know if you have any problems.

 

Regards

Jim

 

Answer 2

at y =2 there are three possible values of x i.e x= 3, 1/2, -2 that would satisfy this cubic function.
 
You can get this values by solving equation 2x^3  - 3x^2 - 11x +8 = 2
 
plug in x=1, -1, 2, -2 and you would find that x=-2 satifies the equation. so x+2 is a factor of 2x^3  - 3x^2 - 11x +6 = 0
 
Now divide 2x^3  - 3x^2 - 11x +6 by x+2 you will get quotient as 2x^2 -7x +3. Now factorise  2x^2 -7x +3 and you will get (2x-1)(x-3)
 
so (x+2)(2x-1)(x-3) are factors of 2x^3  - 3x^2 - 11x +6. Therefore at y=2 there are three possible values of x
i.e. x=-2, 1/2 and 3.
 
I will show you how to find equationof tangent of curve  2x^3  - 3x^2 - 11x +8 at x= -2 rest you can do yourself.
 
To find slope differentiate  y= 2x^3  - 3x^2 - 11x +8 
dy/dx= slope  = 6x^2 - 6x -11
Slope at x = -2
=> 6(-2)^2  - 6(-2) -11 = 25
 
equation of line in slope intercept form is y = mx +b
 
at y=2, x=-2 and m=25
 
2 = 25(-2)  +b
b= 52
 
equation of tangent
 
y = 25x +52
 
:)