Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 2

Use the guidelines of curve sketching to sketch the curve. $y = x^3 + 6x^2 + 9x$

The guidelines of Curve Sketching
A. Domain.
We know that $f(x)$ is a polynomial function having a domain of $(-\infty, \infty)$


B. Intercepts.
Solving for $y$-intercept, when $x=0$.
$y = 0^3 + 6(0)^2 + 9(0) = 0$
Solving for $x$-intercept, when $y = 0$.

$
\begin{equation}
\begin{aligned}
0 &= x^3 + 6x^2 + 9x\\
\\
0 &= x(x^2 + 6x + 9)\\
\\
x &= 0 \text{ and } x^2 + 6x + 9 = 0 \qquad \Longleftarrow \text{(By using Quadratic Formula)}
\end{aligned}
\end{equation}
$

The $x$-intercept are, $x = 0 $ and $x = -3$


C. Symmetry.
The function is not symmetric to both $y$-axis and origin.


D. Asymptotes.
None.


E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, we have $ y' = 3x^2 + 12x + 9$
When $y' =0$, $\quad 0 = 3x^2 + 12x + 9$
The critical numbers are, $x = -1$ and $x = -3$
So, the intervals of increase or decrease are.

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < - 3 & + & \text{increasing on } (-\infty, -3)\\
\hline\\
-3 < x < -1 & - & \text{decreasing on } (-3,-1)\\
\hline\\
x > -1 & + & \text{increasing on } (-1, \infty)\\
\hline
\end{array}
$


F. Local Maximum and Minimum Values.
since $f'(x)$ changes from positive to negative at $x = 3$, then $f(-3) = 0$ is a local maximum. On the other hand, since $f'(x)$ changes from negative to positive of $x = -1$, then $f(-1) = -4$ is a local minimum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= 3x^2 + 12x + 9, \text{ then}\\
\\
f''(x) &= 6x + 12
\end{aligned}
\end{equation}
$

when $f''(x) = 0$, the inflection points is at $x = -2$
Thus, the concavity can be determined by divding the inteval to...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
x < -2 & - & \text{Downward}\\
\hline\\
x > 2 & + & \text{Downard}\\
\hline
\end{array}
$


H. Sketch the Curve