Here are partial solutions to your second midterm exam. It would be in your best interest to take the time to figure out where you made mistakes. As always, if you have questions, please ask.

- Give an example of a polynomial of degree five that has exactly three distinct roots. (5 pts)

Here is a good example: . There’s no need to multiply this out.

- Determine the maximum value of the following function. (5 pts)

The
maximum value of this quadratic is the *y*-coordinate
of the vertex
:

- A farmer has 1200 feet of fencing with which to
enclose 2 adjacent rectangular corrals to keep grazing nuggets from
wandering off. The farmer is planning on building the corrals
right next to a river, so no fence is needed on that side (no, nuggets
can’t swim). See picture below. Find the dimension
*x*and*y*that will maximize the area. (10 pts)

Primary Equation:
,
Secondary Equation:
. Solve for *y* in the Secondary Equation and then
substitute into
the Primary Equation:

Notice that the
resulting equation
is a parabola opening down. To
find the dimensions that maximize the area, we need to find the *x*-coordinate of the vertex (as in #2) and
then plug
this value into any version of the Secondary Equation:

So, the dimensions that maximize the area are .

- Find
__all__of the complex zeros (real and imaginary) of each of the following functions. (10 pts each)

(a)

First, factor out the common factor and then factor the remaining quadratic:

So, are the zeros.

(b)

Factoring by grouping works here. Also, factor the difference of 2 squares:

So, are the zeros.

(c)

The easiest way to do this one is to let and then factor:

So, , which implies that . Therefore, . Because we had to square both sides of an equation to solve, we need to check our solutions. Both do check.

- Find all of the
*x*-intercepts of the following function. (10 pts)

Unfortunately, this
one doesn’t
factor. Shit! The Rational Root Theorem tells us that
the following are __possible__ rational roots:
. Using Descartes Rule of Signs, we see
that there are no positive roots.
So, we don’t need to check any of the positive numbers in the
list. Let’s test
:

Great! So, we
know
is a factor of *f*. By
doing long division, we get:

So,
the *x*-intercepts are
.

- Using the fact that 1 is a zero of the following function with multiplicity 2, write the following polynomial in completely factored form. Include any complex (imaginary) number factors. (10 pts)

Since 1 is a zero with multiplicity 2, is a factor. Doing long division by twice, we get:

Note: You could use factoring by grouping after the first long division.

- Sketch the graph of each of the following
functions. Be sure to label all
*x*and*y*-intercepts and any asymptotes. (10 pts each)

(a)

To find the *x*-intercepts, set equal to 0 and factor
(this function
is quadratic in form):

So, the *x*-intercepts are
. Since the multiplicities are all 1,
the
graph crosses the *x*-axis
at each
of these points. To find the *y*-intercept, plug in
. The *y*-intercept is
. Using these points and the fact that
this is an even degree polynomial (with both “ends” pointing up),
sketch the
graph. This graph has no
asymptotes.

(b)

To find the *x*-intercepts, set the numerator equal to 0
and solve
for *x*. The *x*-intercept
is
. To find the *y*-intercept, plug in
. The *y*-intercept is
. The vertical asymptote is
since this is the value that makes the
denominator 0. By comparing the
degrees of the numerator and denominator (degrees are equal, so we get
the
ratio of the leading coefficients), we see that the horizontal
asymptote is
. Use this information to sketch the
graph.

- Solve the following inequality. Write your answer using intervals. (10 pts)

You must get common denominators (or keep track of the fact that ):

The *x*-intercepts of the corresponding function
are
,
and the vertical asymptote is
. These 3 points divide the *x*-axis into 4 intervals. By picking
test points, we can determine where the function
is above and below the *x*-axis.
The solution is
. Note: 0 is not included because it is
a
vertical asymptote.