PreCalculus Honors

Unit Test #2

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Part 1: Calculator Allowed

 

For questions 1-3, let f x =3 x 4 x 3 9 x 2 +3x.

 

1.

Locate intervals where f x is increasing and decreasing.

(5)

An Image

 

The graph is decreasing on x ,1.1887 0.1651,1.2736 and increasing on x 1.1887,0.1651 1.2736, .

 

2.

Locate and classify the extrema of f x .

(4)

There are minima at x=1.1887 and x=1.2736. There is a maximum at x=0.1651.

 

3.

Locate all real zeros of f x .

(4)

I won’t re-graph or re-label…but they are at -1.7321, 0, 0.3333 and 1.7321.

 

4.

Find the quotient and remainder: x 5 +2 x 4 6 x 3 + x 2 5x+1 x 3 +1

(5)

x 2 +2x 6 x 3 +1 x 5 +2 x 4 6 x 3 + x 2 5x +1 x 5 0 x 4 0 x 3 + x 2 2 x 4 6 x 3 +0 x 2 5x +1 2 x 4 +0 x 3 +0 x 2 +2x 6 x 3 +0 x 2 7x +1 6 x 3 0 x 2 0x 6 7x +7

The quotient is x 2 +2x6 and the remainder is 7x+7.

PreCalculus Honors

Unit Test #2

 

Part 2: No Calculator Allowed

 

5.

Use the remainder theorem to find g 2 if g x = x 3 2 x 2 +8x4.

(4)

Let’s use synthetic division.

2 1 2 8 4 2 0 16 1 0 8 12

Thus, g 2 =12.

 

6.

List the possible rational roots of f x =2 x 5 x 4 10 x 3 +5 x 2 +12x6.

(4)

These will be factors of 6 over factors of 2. ± 1,2,3,6, 1 2 , 3 2 .

 

7.

Write a polynomial that has the following zeros and multiplicities:

3, 2i, 1 (multiplicity 3)

(5)

Don’t forget that complex non-real zeros must come in conjugate pairs. x3 x 2i x 2+i x+1 3

 

8.

List all complex zeros (and multiplicities) of h x = x 2 4 x4 2 x 2 +1 .

(6)

x 2 4x=±2; x4 2 x=4 with multiplicity 2; x 2 +1x=±i.

 

9.

Find all complex zeros of p x = x 4 3 x 3 x 2 +13x10.

(7)

The possible rational roots are ± 1,2,5,10 . I notice that p 1 =131+1310=0, so 1 is a root.

1 1 3 1 13 10 1 2 3 10 1 2 3 10 0

That makes the factorization x1 x 3 2 x 2 3x+10 . Clearly, neither 1 nor negative 1 will work for the remaining piece…how about 2? p 2 = 2 3 2 2 2 32+10=886+100. Hmmm…maybe negative 2? p 2 = 2 3 2 2 2 3 2 +10=88+6+10=0, so 2 is a root.

2 1 2 3 10 2 8 10 1 4 5 0

This makes the factorization x1 x+2 x 2 4x+5 . Time for the quadratic formula!

x= 4± 4 2 4 1 5 2 1 = 4± 1620 2 = 4± 4 2 = 4±2i 2 =2±i

Thus (finally) we have the list of complex zeros: 1,2,2±i.

 

10.

Determine the end behavior of q x =2 x 5 x 4 +3 x 3 x 2 x+1.

(3)

Since the degree is odd, the ends will point in opposite directions. Since the leading coefficient is negative, the right side will point down. Thus, the end behavior is +, .


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