# How do you use the first and second derivatives to sketch y=(x^3)-(6x^2)+5x+12?

Jun 28, 2016

The first derivative allows you to determine the location of the critical points of the function. The second allows you to determine the nature of these points.

#### Explanation:

The first derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ describes how y changes with x, ie a gradient. The second derivative $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ effectively describes how the gradient changes with x, ie the curvature. At a turning point, the gradient will be zero, so the turning points can be located by setting the first derivative equal to zero.

$y ' = 3 {x}^{2} - 12 x + 5 = 0$

Using the quadratic formula:

$x = \frac{12 \pm \sqrt{144 - 60}}{6}$

$x = 3.528 \mathmr{and} x = 0.472$

This gives the location of the turning points, we now find the second derivative to determine their natures.

$y ' ' = 6 x - 12$

We can see that for x = 3.528 that y'' > 0, x = 0.472 that y'' < 0.
For positive curvature, it will be curving upwards (think a positive, happy smiley face :)) so the point there will be in a trough, ie a minimum. The point with negative curvature will be curving downwards so will be a maximum.

With this, plug in the x values to find the values of the function at these critical points, mark them taking note of their natures and then, taking into account other useful features like intercepts and asymptotes, plot the graph.