Question

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

Answer 1

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]}