# How do you find the critical points for #y = -0.1x^2 − 0.5x + 15#?

##### 1 Answer

#### Answer:

You calculate the function's first derivative and make it equal to zero.

#### Explanation:

A function's *critical points* are points in the function's domain at which the first derivative is either **equal to zero** or **undefined**.

For afunction

#f^'(c) = 0" "# or#" "f^'(c) = "undefined"#

So, the first thing you need to do is find the function's first derivative

#d/dx(y) = d/dx(-0.1x^2 - 0.5x + 15)#

#y^' = -0.2x - 0.5#

There are no values of

#y^'(x) = 0#

#-0.2x - 0.5 = 0#

#-0.2x = 0.5 implies x= 0.5/((-0.2)) = -2.5#

The function

The point of the graph will correspond to

#y(2.5) = -0.1 * (-2.5)^2 - 0.5 * (-2.5) + 15#

#y(2.5) = 15.625#

graph{-0.1x^2 - 0.5x + 15 [-41.1, 41.1, -20.56, 20.56]}