# How do you sketch the curve y=x^3-3x^2-9x+5 by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Nov 29, 2016

graph{x^3-3x^2-9x+5 [-145.9, 172.6, -85.6, 73.6]} FIrst determine the interval of definition, then the behavior of first and second derivatives and the behavior of the function as $x$approaches $\pm \infty$

#### Explanation:

1) The function is a polynomial and is defined for $x \in \left(- \infty , + \infty\right)$

2) The highest monomial is of odd order so:

${\lim}_{x \to - \infty} y \left(x\right) = - \infty$
${\lim}_{x \to + \infty} y \left(x\right) = + \infty$

Also $\frac{y \left(x\right)}{x}$ does not have a finite limit, so the function has no asymptotes.

3) Calculate the first and second derivative:

$y ' \left(x\right) = 3 {x}^{2} - 6 x - 9$
$y ' ' \left(x\right) = 6 \left(x - 1\right)$

The points where $y ' \left(x\right) = 0$ are:

$x = \frac{3 \pm \sqrt{9 + 27}}{3} = \frac{3 \pm 6}{3}$

${x}_{1} = - 1 , {x}_{2} = 3$

In both points the second derivative is non null, so these are local extrema and not inflection points.

If we analyze the sign of $y ' \left(x\right)$ we see that:

$y ' \left(x\right) > 0$ for $x \in \left(- \infty , - 1\right) \mathmr{and} x \in \left(3 , + \infty\right)$
$y ' \left(x\right) < 0$ for $x \in \left(- 1 , 3\right)$

So, $y \left(x\right)$ starts from $- \infty$, grows util $x = - 1$ where it reaches a local maximum, decreases until $x = 3$ where it reaches a local minimum, that starts growing again.

4) Analyze the value of the function in the local extrema:

$y {\left(x\right)}_{x = - 1} = {\left(- 1\right)}^{3} - 3 {\left(- 1\right)}^{2} - 9 \left(- 1\right) + 5 = - 1 - 3 + 9 + 5 = 10$

$y {\left(x\right)}_{x = 3} = {3}^{3} - 3 \cdot {3}^{2} - 9 \cdot 3 + 5 = 27 - 27 - 27 + 5 = - 22$

So there will be three real roots: one in $\left(- \infty , - 1\right)$, one in $\left(- 1 , 3\right)$ and one in $\left(3 , + \infty\right)$