Question

question 5
4 pts
find the points on the curve y = 2x^3 + 3x^2 - 12x + 9 where the tangent line is horizontal.
(-2,29) and (1,2)
(-3,29) and (2,1)
(2,1) and (29,-3)
the correct answer is not listed.
(-2,1) and (29,2)

Explanation:

Step1: Find the derivative

The derivative of $y = 2x^{3}+3x^{2}-12x + 9$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=6x^{2}+6x-12$.

Step2: Set the derivative equal to 0

Since the slope of a horizontal tangent line is 0, we set $y' = 0$. So, $6x^{2}+6x - 12=0$. Divide through by 6 to get $x^{2}+x - 2=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}+x - 2=(x + 2)(x - 1)=0$. Then, $x=-2$ or $x = 1$.

Step4: Find the y - values

When $x=-2$, $y=2(-2)^{3}+3(-2)^{2}-12(-2)+9=2(-8)+3(4)+24 + 9=-16 + 12+24 + 9=29$.
When $x = 1$, $y=2(1)^{3}+3(1)^{2}-12(1)+9=2 + 3-12 + 9=2$.

Answer:

A. (-2,29) and (1,2)