# How do you determine the intervals for which the function is increasing or decreasing given f(x)=(x+3)^2-4?

Jan 26, 2018

Since the second derivative is greater than 0 (positive), -3 is a local Minimum

#### Explanation:

Second Derivative Test

When a function's slope is zero at x, and the second derivative at x is:

less than 0, it is a local maximum
greater than 0, it is a local minimum

$y = f \left(x\right) = {\left(x + 3\right)}^{2} - 4$

$y ' = \left(\frac{d}{\mathrm{dx}}\right) y = 2 \cdot \left(x + 3\right)$

Equating to 0, we get, color(red)(x = -3)

Second Derivative Test :

y" = ((d^2 y)/dx^2) = 2#

Since the second derivative is greater than 0 (positive), -3 is a local Minimum