SOLUTIONS FOR TEST#3

 

The problems are not numbered because of the different forms of the test. I have labeled each section by the bold title or the introduction to the problem.

 

Solve each equation. Simplify any fractions, if possible.

a.  (2y + 5)(3y – 1)(y + 8) = 0    θuse the Zero Products Principle and set each = 0

                                                    and solve

2y + 5 = 0            3y – 1 = 0            y + 8 = 0

y =                   y =                   y = 8

 

b.    

Usually we need to check for extraneous roots because of the squaring, but here we only get one, and it gives a correct zero for the equation.

c. 

 

The key here is to use the meaning of the absolute value—the distance away from zero. Remember the absolute value of a negative distance is undefined because there is no such thing. So, the definition tells us the distance can be located on either side of zero, thus, the need for the positive and negative values, once the absolute value is isolated. Using the definition in this way removes the symbols.

d. 

 

The key to this problem is getting the common denominator AND multiplying every term of the equation by the common denominator. Then the denominators divide out and leave a fairly simple equation to solve.

 

 

Maximizing Area. Do not solve the problem but answer: why do you think the stonemason used the wall of the house as one side of the enclosure?

 

There are various answers to this, but points to bring out are 1) using one wall of the house leaves materials used for 3 walls of the enclosure to make it larger or 2) it is there and long enough for the materials available.

 

Given B(x) = x 5 + 5x 2 – 6 with its graph below.

a.  max F(– 1.28, – 1.24)                   min G(– 0.01, – 6)

 

  1. increasing: (

 

decreasing: [– 1.28, – 0.01]

 

 

Given c(x) = x 2 – 14x + 7.

 

a.  c(x) = ( x 2 – 14x +   ) + 7        

     c(x) = (x 2 – 14x + 49   ) + 7 – 49

     c(x) = ( x – 7) 2 – 42    

 

So, the vertex is ( 7, – 42)              

To complete the square, take half of the middle term coefficient, square it and add it to the trinomial in parentheses. Remember to subtract the same amount outside the parentheses so the net amount added is 0.

 

  1. The minimum value is c(x) = – 42 because the leading term of the parabola is positive. Remember the minimum/maximum value of the function that is a parabola is the y – value of the vertex.
  2. The number of real zeros of c(x) is 2. To justify this statement you could use one of two methods.
    1. Graph the function and see that it crosses the x – axis in two places. BUT to justify this, sketch the graph that you see on the calculator.
    2. Use the discriminate to find the number of zeros of the function.

  b 2 – 4ac = (– 14) 2 – 4(1)(7) = 196 – 28 > 0

 

Remember there are 3 different possibilities on the number of zeros:

   Discriminant > 0 θ 2 real

   Discriminant = 0 θ 1 real

   Discriminant < 0 θ none

 

Complete the following table about polynomial functions. Answers are given row by row with the order: name, degree, leading term.

 

Cubic, 3, – 12x 3            Many of you only gave the leading coefficient vs. the full term.

Linear, 1, 0.55x

Quintic, 5, 6t 5     Several of you gave the first term given vs. the highest degree term.

 

 

Given P(x) = x 3 + 4x 2 + x – 6, with x – values 1, – 2, – 3 as possible zeros of the function.

 

  1. synthetic division, long division, substitution, sign change using Intermediate Value Theorem
  2. using synthetic division

 

1 

     1  5   6   0

  1. The remainder must be zero.
  2. Long division

 

  1. Here is the synthetic substitution for      (Note: it is difficult to line up numbers in the grid when typed.)

 

P(9) =   9   

                  1  13 118  1056               

 

 

 

Given R(x) = (x + 3) 2 (x – 1) 2. Write answers in complete sentences.

 

  1. The zeros are x = – 3 and x = 1. This can be seen by setting each factor = 0 and solving.
  2. – 3 has multiplicity of 2 and 1 has multiplicity of 2.
  3. The degree of the function is 4 because the first part of the expression (x + 3) 2 yields x 2 with other terms and so does the second part of the expression. So, (x 2)( x 2) = x 4, which gives the degree.
  4. This part of the question asks you to identify the correct graph of the function R(x). However, just saying it is #4 is not nearly enough. The function can be graphed on your calculator and seen what it looks like. Therefore, the meat of this question is understanding and identifying the characteristics that justify the graph #4.

 

Since – 3 and 1 are the zeros of the function, we know the graph crosses or touches the x – axis at these points. Because the multiplicity of each zero is even, we know that the graph should be tangent to the x – axis at the points. The degree of the function is even, so it acts like a parabola using the sign of the leading term to determine if it opens up or down. The sign is positive, so the graph opens up.