# How do you sketch the graph y=x^3+2x^2+x using the first and second derivatives?

Jan 26, 2017

Refer Explanation section

#### Explanation:

Given -

$y = {x}^{3} + 2 {x}^{2} + x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + 4 x + 1$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 x + 4$

$\frac{\mathrm{dx}}{\mathrm{dy}} = 0 \implies 3 {x}^{2} + 4 x + 1$

$x = \frac{\left(- b\right) \pm \sqrt{{b}^{2} - \left(4 \cdot a \cdot c\right)}}{2 a}$

$x = \frac{\left(- 4\right) \pm \sqrt{{4}^{2} - \left(4 \cdot 3 \cdot 1\right)}}{2 \cdot 3}$

$x = \frac{\left(- 4\right) \pm \sqrt{16 - \left(12\right)}}{6}$

$x = \frac{\left(- 4\right) \pm \sqrt{16 - 12}}{6}$

$x = \frac{\left(- 4\right) \pm \sqrt{16 - 12}}{6}$

$x = \frac{\left(- 4\right) \pm \sqrt{4}}{6}$

$x = \frac{\left(- 4\right) \pm 2}{6}$

$x = \frac{- 4 - 2}{6} = \frac{- 6}{6} = - 1$

$x = \frac{- 4 + 2}{6} = \frac{- 2}{6} = - \frac{1}{3}$

At $x = - 1$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 \left(- 1\right) + 4 = - 6 + 4 = - 2$

At $x = - 1$

dy/dx=0; (d^2y)/(dx^2)<0

Hence the function has a maximum.
At $x = - 1$ the curve is concave downwards.

At $x = - \frac{1}{3}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 \left(- \frac{1}{3}\right) + 4 = - 2 + 4 = 2$

At $x = - 1$

dy/dx=0; (d^2y)/(dx^2)>0

Hence the function has a minimum.
At $x = - 1$ the curve is concave upwards.

graph{x^3+2x^2+x [-10, 10, -5, 5]}

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