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

1 Answer
Aug 13, 2015

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(x)#, a critical point #c# will thus satisfy

#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 #x# for which #y^'(x)# is undefined, so look for values of #x# that make #y^'(x) = 0#.

#y^'(x) = 0#

#-0.2x - 0.5 = 0#

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

The function #y = -0.1x^2 - 0.5x + 15# will have a critical point at #x = -2.5#.

The point of the graph will correspond to #(-2.5, y(-2.5))#

#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]}