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

Aug 13, 2015

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 $f \left(x\right)$, a critical point $c$ will thus satisfy

${f}^{'} \left(c\right) = 0 \text{ }$ or $\text{ "f^'(c) = "undefined}$

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

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left(- 0.1 {x}^{2} - 0.5 x + 15\right)$

${y}^{'} = - 0.2 x - 0.5$

There are no values of $x$ for which ${y}^{'} \left(x\right)$ is undefined, so look for values of $x$ that make ${y}^{'} \left(x\right) = 0$.

${y}^{'} \left(x\right) = 0$

$- 0.2 x - 0.5 = 0$

$- 0.2 x = 0.5 \implies x = \frac{0.5}{\left(- 0.2\right)} = - 2.5$

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

The point of the graph will correspond to $\left(- 2.5 , y \left(- 2.5\right)\right)$

$y \left(2.5\right) = - 0.1 \cdot {\left(- 2.5\right)}^{2} - 0.5 \cdot \left(- 2.5\right) + 15$

$y \left(2.5\right) = 15.625$

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