1. First, we use the leading-term test to determine the end behavior of the graph. The leading term is $-2x^4$. The degree, 4, is even, and the coefficient, −2, is negative. Thus the end behavior of the graph as $x\to \infty$ and as $x\to -\infty$ can be sketched as follows.

   

2. The zeros of the function are the first coordinates of the x-intercepts of the graph. To find the zeros, we solve $h(x)=0$ by factoring and using the principle of zero products.

   $$\begin{array}{l}\begin{array}{rcll}-2x^4+3x^3& =& 0& \\ -x^3(2x-3)& =& 0& \mathbf{\text{Factoring}}\end{array}\\ \begin{array}{rclcrcll}-x^3& =& 0& \mathit{\text{or}}& 2x-3& =& 0& \mathbf{\text{Using the principle of zero products}}\\ x& =& 0& \mathit{\text{or}}& x& =& \frac{3}{2}.& \end{array}\end{array}$$

   The zeros of the function are 0 and $\frac{3}{2}$. Note that the multiplicity of 0 is 3 and the multiplicity of $\frac{3}{2}$ is 1. The x-intercepts are (0, 0) and $\left(\frac{3}{2},0\right)$.

3. The zeros divide the x-axis into three intervals:

   $$(-\infty ,0),(0,\frac{3}{2}),\text{and}(\frac{3}{2},\infty ).$$

   

   The sign of $h(x)$ is the same for all values of x in each of the three intervals. That is, $\mathit{h}\mathbf{(}\mathit{x}\mathbf{)}$ is positive for all x-values in an interval or $\mathit{h}\mathbf{(}\mathit{x}\mathbf{)}$ is negative for all x-values in an interval. To determine which, we choose a test value for x from each interval and find $h(x)$.

   Interval	$(-\infty ,0)$	$\left(0,\frac{3}{2}\right)$	$\left(\frac{3}{2},\infty \right)$
   Test Value, x	−1	1	2
   Function Value, $\mathit{h}\mathbf{(}\mathit{x}\mathbf{)}$	−5	1	−8
   Sign of $\mathit{h}\mathbf{(}\mathit{x}\mathbf{)}$	−	$+$	−
   Location of Points on Graph	Below x-axis	Above x-axis	Below x-axis

   

   This test-point procedure also gives us three points to plot. In this case, we have (−1, −5), (1, 1), and (2, −8).

4. To determine the y-intercept, we find $h(0):$

   $$\begin{array}{rcl}h(x)& =& -2x^4+3x^3\\ h(0)& =& -2\cdot 0^4+3\cdot 0^3=0.\end{array}$$

   The y-intercept is (0, 0).

5. A few additional points are helpful when completing the graph.

   x	$h(x)$
   −1.5	−20.25
   −0.5	−0.5
   0.5	0.25
   2.5	−31.25

   

6. The degree of h is 4. The graph of h can have at most 4 x-intercepts and at most 3 turning points. In fact, it has 2 x-intercepts and 1 turning point. The zeros, 0 and $\frac{3}{2}$, each have odd multiplicities: 3 for 0 and 1 for $\frac{3}{2}$. Since the multiplicities are odd, the graph crosses the x-axis at 0 and $\frac{3}{2}$. The end behavior of the graph is what we described in step (1). As $x\to \infty$ and also as $x\to -\infty ,h(x)-\infty$. The graph appears to be correct.

1. The leading term is $2x^3$. The degree, 3, is odd, and the coefficient, 2, is positive. Thus the end behavior of the graph will appear as follows.

   

2. To find the zeros, we solve $f(x)=0$. Here we can use factoring by grouping.

   $$\begin{array}{rcll}2x^3+x^2-8x-4& =& 0& \\ x^2(2x+1)-4(2x+1)& =& 0& \mathbf{\text{Factoring by grouping}}\\ (2x+1)(x^2-4)& =& 0& \\ (2x+1)(x+2)(x-2)& =& 0& \mathbf{\text{Factoring a difference of squares}}\end{array}$$

   The zeros are $-\frac{1}{2},-2$, and 2. Each is of multiplicity 1. The x-intercepts are $(-2,0),\left(-\frac{1}{2},0\right)$, and (2, 0).

3. The zeros divide the x-axis into four intervals:

   $$(-\infty ,-2),(-2,-\frac{1}{2}),(-\frac{1}{2},2),\text{and}(2,\infty ).$$

   

   We choose a test value for x from each interval and find $f(x)$.

   Interval	$(-\infty ,-2)$	$\left(-2,-\frac{1}{2}\right)$	$\left(-\frac{1}{2},2\right)$	$(2,\infty )$
   Test Value, x	−3	−1	1	3
   Function Value, $\mathit{f}\mathbf{(}\mathit{x}\mathbf{)}$	−25	3	−9	35
   Sign of $\mathit{f}\mathbf{(}\mathit{x}\mathbf{)}$	−	$+$	−	$+$
   Location of Points on Graph	Below x-axis	Above x-axis	Below x-axis	Above x-axis

   

   The test values and corresponding function values also give us four points on the graph: (−3, −25), (−1, 3), (1, −9), and (3, 35).

4. To determine the y-intercept, we find $f(0):$

   $$\begin{array}{rcl}f(x)& =& 2x^3+x^2-8x-4\\ f(0)& =& 2\cdot 0^3+0^2-8\cdot 0-4=-4.\end{array}$$

   The y-intercept is (0, −4).

5. We find a few additional points and complete the graph.

   x	$f(x)$
   −2.5	−9
   −1.5	3.5
   0.5	−7.5
   1.5	−7

   

6. The degree of f is 3. The graph of f can have at most 3 x-intercepts and at most 2 turning points. It has 3 x-intercepts and 2 turning points. Each zero has a multiplicity of 1; thus the graph crosses the x-axis at −2, $-\frac{1}{2}$, and 2. The graph has the end behavior described in step (1). As $x\to -\infty ,h(x)\to -\infty$, and as $x\to \infty \text{,}h(x)\to \infty$. The graph appears to be correct.

1. The leading term is $x^4$. The degree, 4, is even, and the coefficient, 1, is positive. The sketch below shows the end behavior.

   

2. To find the zeros, we solve $g(x)=0:$

   $$(x+1){(x-2)}^2(x-4)=0.$$

   The zeros are −1, 2, and 4; 2 is of multiplicity 2; the others are of multiplicity 1. The x-intercepts are (−1, 0), (2, 0), and (4, 0).

3. The zeros divide the x-axis into four intervals:

   $$(-\infty ,-1),(-1,2),(2,4),\text{and}(4,\infty ).$$

   

   We choose a test value for x from each interval and find $g(x)$.

   Interval	$(-\infty ,-1)$	(−1, 2)	(2, 4)	$(4,\infty )$
   Test Value, x	−1.25	1	3	4.25
   Function Value, $\mathit{g}\mathbf{(}\mathit{x}\mathbf{)}$	$\approx 13.9$	−6	−4	$\approx 6.6$
   Sign of $\mathit{g}\mathbf{(}\mathit{x}\mathbf{)}$	$+$	−	−	$+$
   Location of Points on Graph	Above x-axis	Below x-axis	Below x-axis	Above x-axis

   

   The test values and the corresponding function values also give us four points on the graph: (−1.25, 13.9), (1, −6), (3, −4), and (4.25, 6.6).

4. To determine the y-intercept, we find $g(0):$

   $$\begin{array}{rcl}g(x)& =& x^4-7x^3+12x^2+4x-16\\ g(0)& =& 0^4-7\cdot 0^3+12\cdot 0^2+4\cdot 0-16=-16.\end{array}$$

   The y-intercept is (0, −16).

5. We find a few additional points and draw the graph.

   x	$g(x)$
   −0.5	−14.1
   0.5	−11.8
   1.5	−1.6
   2.5	−1.3
   3.5	−5.1

   

6. The degree of g is 4. The graph of g can have at most 4 x-intercepts and at most 3 turning points. It has 3 x-intercepts and 3 turning points. One of the zeros, 2, has a multiplicity of 2, so the graph is tangent to the x-axis at 2. The other zeros, −1 and 4, each have a multiplicity of 1 so the graph crosses the x-axis at −1 and 4. The graph has the end behavior described in step (1). As $x\to \infty$ and as $x\to -\infty ,g(x)\to \infty$. The graph appears to be correct.

We find $f(a)$ and $f(b)$ or $g(a)$ and $g(b)$ and determine whether they differ in sign. The graphs of $f(x)$ and $g(x)$ provide visual checks of the conclusions.

1. $f(-4)={(-4)}^3+{(-4)}^2-6(-4)=-24$,

   $f(-2)={(-2)}^3+{(-2)}^2-6(-2)=8$

   Note that $f(-4)$ is negative and $f(-2)$ is positive. By the intermediate value theorem, since $f(-4)$ and $f(-2)$ have opposite signs, then $f(x)$ has at least one zero between −4 and −2. The following graph confirms this.

   

2. $f(-1)={(-1)}^3+{(-1)}^2-6(-1)=6$,

   $f(3)=3^3+3^2-6(3)=18$

   Both $f(-1)$ and $f(3)$ are positive. Thus the intermediate value theorem does not allow us to determine whether there is a real zero between −1 and 3. Note that the graph of $f(x)$ above shows that there are two zeros between −1 and 3.

3. $g\left(-\frac{1}{2}\right)=\frac{1}{3}{\left(-\frac{1}{2}\right)}^4-{\left(-\frac{1}{2}\right)}^3=\frac{7}{48}$,

   $g\left(\frac{1}{2}\right)=\frac{1}{3}{\left(\frac{1}{2}\right)}^4-{\left(\frac{1}{2}\right)}^3=-\frac{5}{48}$

   Since $g\left(-\frac{1}{2}\right)$ and $g\left(\frac{1}{2}\right)$ have opposite signs, $g(x)$ has at least one zero between $-\frac{1}{2}$ and $\frac{1}{2}$. The following graph confirms this.

   

4. $g(1)=\frac{1}{3}{(1)}^4-1^3=-\frac{2}{3}$,

   $g(2)=\frac{1}{3}{(2)}^4-2^3=-\frac{8}{3}$

   Both $g(1)$ and $g(2)$ are negative. Thus the intermediate value theorem does not allow us to determine whether there is a real zero between 1 and 2. Note that the graph of $g(x)$ above shows that there are no zeros between 1 and 2.