The $y$-intercept is $(0,\answer [given]{0})$. Place this point on your plot.

Which of the following are vertical asymptotes? Select all that apply.

In this case, $f(x) =2x^3-3x^2-12x$, we can find the $x$-intercepts. There are three $x$ intercepts. Call them $a$, $b$, and $c$, and order them such that $a<b<c$. Then

Which of the following best describes the end behavior of $f$ as $x \to \infty$?

Compute $f'(x)$ and $f''(x)$,

Mark the critical points $x=2$ and $x=-1$ on your plot.

Similarly, since for $-1<x<2$, $(x+1)>0$ and $(x-2)<0$, the derivative is negative there, and, therefore, our function is decreasing on $(-1,2)$.

And, for $x>2$, both factors $(x+1)$ and $(x-2)$ are positive, and, therefore, our function is increasing on $(2,\infty )$.

Hence $x=-1$, corresponding to the point $(-1,7)$ is a local maximum and $x=2$, corresponding to the point $(2,-20)$ is local minimum of $f(x)$. Identify this on your plot.

In order to locate inflection points, we have to solve the equation $f''(x) = 0$, thus we need to solve $12\left (x-\frac {1}{2}\right )=0$ for $x$.

The solution to this is $x = \answer {\frac {1}{2}}$.

This is only a possible inflection point; we still have to check whether concavity changes there.

We have that $f''(x)<0$ for $x<\frac {1}{2}$, therefore $f$ is concave downon $\left (-\infty ,\frac {1}{2}\right )$.

Similarly, $f''(x)>0$ for $x>\frac {1}{2}$, therefore $f$ is concave up on $\left (\frac {1}{2},\infty \right )$.

So, concavity changes at $x=\frac {1}{2}$, and therefore, this point is a point of inflection.

Since all of this behavior as described above occurs on the interval $[-2,4]$, we now have a complete sketch of $f(x)$ on this interval, see the figure below.

Since this function is piecewise defined, we will analyze the the graph of $f$ on intervals $(-\infty ,0)$ and $[0,\infty )$ separately.

Because $f$ is piecewise defined, and potentially discontinuous at $0$, it is important to understand the behavior of $f$ near $x = 0$.

$$\lim _{x \to 0^-} f(x) = \answer [given]{2}$$

$$\lim _{x \to 0^+} f(x) = \answer [given]{3}$$

Moreover,

$$f(0) = \answer [given]{3}$$

Record this information on our graph with filled and unfilled circles.

Which of the following are vertical asymptotes on $(-\infty ,0)$? Select all that apply.

Which of the following are vertical asymptotes on $(0,\infty )$? Select all that apply.

Which of the following best describes the end behavior of $f$ as $x \to \infty$?

Which of the following best describes the end behavior of $f$ as $x \to -\infty$?

We mark the location of the horizontal asymptote:

The derivative of $f$ on $(-\infty , 0)$ is $$f'(x)= \answer [given]{xe^x+e^x}=e^x(\answer [given]{x+1})$$

The derivative of $f$ on $(0, \infty )$ is $$f'(x)= \answer [given]{4x^3-2x}=4x\left (\answer [given]{x^2-\frac {1}{2}}\right )$$

The critical points are the points where $f'(x) = 0$ or $f'(x)$ does not exist. $0$ is a critical point, since we have already seen it is a point of discontinuity for $f$, and thus $f'(0)$ does not exist there. On the interval $(-\infty , 0)$, $f$ has a critical point at $x = \answer [given]{-1}$. On the interval $(0,\infty )$, $f$ has a critical point at $x = \answer [given]{\frac {1}{\sqrt {2}}}$. Mark the critical points $x=-1$ and $x=\frac {1}{\sqrt {2}}$ on your plot.

Does the function $f$ have a critical point at $x=-\frac {1}{\sqrt {2}}$? Explain.

On the interval $(-\infty , -1)$, the derivative $f'(x)=e^x(x+1)$ and has the same sign as the factor $(x+1)$, which is negative.

Therefore, $f$ is decreasing on $(-\infty , -1)$.

On the interval $(-1, 0)$, the derivative $f'(x)$ has the same sign as the factor $(x+1)$, which is positive.

Therefore, $f$ is increasing on $(-1, 0)$.

On the interval $\left (0, \frac {1}{\sqrt {2}}\right )$, the derivative $f'(x)$ does not change the sign.

Since $f'\left (\frac {1}{4}\right )= \frac {1}{16}-\frac {1}{2}<0$, it follows that $f$ is decreasing on $\left (0, \frac {1}{\sqrt {2}}\right )$.

On the interval $\left (\frac {1}{\sqrt {2}}, \infty \right )$, the derivative $f'(x)$ does not change the sign. Since $f'(1)= 4\left (1-\frac {1}{2}\right )>0$, $f$ is increasing on $\left (\frac {1}{\sqrt {2}}, \infty \right )$.

The second derivative of $f$ on $(0, \infty )$ is $$f''(x)= \answer [given]{12x^2-2}=12\left (\answer [given]{x^2-\frac {1}{6}}\right )$$ The candidates for the inflection points are points where $f''(x) = 0$.

On $(-\infty , 0)$, $f''$ has one zero, namely $x = \answer [given]{-2}$. The sign of $f''$ changes from negative to positive] at this point.

On $(0, \infty )$, $f''$ has one zero, namely $x = \answer [given]{\frac {1}{\sqrt {6}}}$. The sign of $f''$ changes from negative to positive] through this point.

Since all of this behavior as described above occurs on the interval $[-5,2]$, we now have a complete sketch of $y=f(x)$ on this interval, see the figure below.