Question

Find the equations of any horizontal tangents to the curve
y = x^3 - 3x ^2 - 9x + 2

Answer 1

Answers:

y = -25

y = 7

Step-by-step explanation:

Find the derivative of the function.

$\frac{dy}{dx} y = 3x^2-6x-9$  

The zeros of the derivative are the local extrema (Maximum & Minimum)

$0 = 3x^2-6x-9$

Take out the greatest common factor, 3, to get this equation:

$0 = 3(x^2-2x-3)$

Factor out the brackets.

$0 = 3(x - 3) ( x + 1 )$

You are left with two possible values for x:

x = 3

x = -1

By substituting each value into the original function $y = x^3 - 3x^2 - 9x + 2$, you get the equation of any horizontal tangents to the curve.

When x = 3:

$y = (3)^3 - 3(3)^2 - 9(3) + 2 = -25$       Meaning y = -25

When x = -1:

$y = (-1)^3 - 3(-1)^2 - 9(-1) + 2 = 7$    Meaning  y = 7