Math 1150: Precalculus, Instructor: Dana Ernst

Solutions for Midterm Exam  #2

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.

 

1. 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.

 

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

                         

 

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

                         

 

3. 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 .

 

4. 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.

 

5. 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 .         

 

6. 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.

 

7. 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.

 

8. 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.