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

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Answer 1 · 2015-08-13T23:32:10

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

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]}

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