1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. (1 point)
4, -14, and 5 + 8i


a) f(x) = x4 - 362.5x2 + 1450x - 4984
b) f(x) = x4 - 9x3 + 32x2 - 725x + 4984
c) f(x) = x4 - 67x2 + 1450x - 4984
d) f(x) = x4 - 9x3 - 32x2 + 725x - 4984

2. State how many imaginary and real zeros the function has. (1 point)
f(x) = x5 + 7x4 + 2x3 + 14x2 + x + 7


a) 3 imaginary; 2 real
b) 4 imaginary; 1 real
c) 0 imaginary; 5 real
d) 2 imaginary; 3 real

3. Write a linear factorization of the function.(1 point)
f(x) = x4 + 16x2


a) f(x) = x2 (4x + i)(4x - i)
b) f(x) = x2 (x + 4i)2
c) f(x) = x2 (x + 4i)(x - 4i)
d) f(x) = x2 (4x + i)2

4. Using the given zero, find one other zero of f(x). (1 point)
i is a zero of f(x).= x4 - 2x3 + 38x2 - 2x + 37.


a) -1 + i
b) -i
c) -1 - i
d) 1

5. State the domain of the rational function. (1 point)
f(x) = seven divided by quantity fourteen minus x.


a) All real numbers except 14
b) All real numbers except -14 and 14
c) All real numbers except 7
d) All real numbers except -7 and 7

6. State the vertical asymptote of the rational function.(1 point)
f(x) = quantity x minus seven times quantity x plus four divided by quantity x squared minus four.


a) x = 2, x = -2
b) None
c) x = 7, x = -4
d)x = -7, x = 4

7. State the horizontal asymptote of the rational function. (1 point)
f(x) = quantity x plus nine divided by quantity x squared plus eight x plus five.


a) y = x
b) y = 0
c) y = 9
d) None

8. State the horizontal asymptote of the rational function. (1 point)
f(x) = quantity x squared plus nine x minus nine divided by quantity x minus nine.


a) y = 3
b) None
c) y = 9
d) y = -9

9. Give an example of a rational function that has a horizontal asymptote of y = 2/9.

Respuesta :

Question 1:

Given the three zeros
[tex]x = 4[/tex]
[tex]x = -14[/tex]
[tex]x = 5+8i[/tex]
The other zero will be the conjugate pair of [tex]5+8i[/tex]
[tex]x = 5-8i[/tex]

Writing these zeros in factorized form of f(x) gives:
[tex](x-4)(x+14)(x-(5+8i))(x-(5-8i))[/tex] → Multipy out the first two brackets to get
[tex](x^2+10x-56)(x-5-8i)(x-5+8i)[/tex]
[tex](x^2+10x-56)(x^2-5x+8ix-5x+25-40i-8ix+40i-64i^2)[/tex]
[tex](x^2+10x-56)(x^2-10x+25-64(-1))[/tex]
[tex](x^2+10x-56)(x^2-10x+25+64)[/tex]
[tex](x^2+10x-56)(x^2-10x+89)[/tex]→ Multiply out further
[tex]x^4-10x^3+89x^2+10x^3-100x^2+890x-56x^2+560x-4984[/tex]
[tex]x^4+89x^2-100x^2-56x^2+1450x-4984[/tex]
[tex]x^4-67x^2+1450x-4984[/tex]

Answer: C
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Question 2

Given [tex]x^5+7x^4+2x^3+14x^2+x+7[/tex]
Collecting two terms together to get
[tex](x^5+7x^4)+(2x^3+14x^2)+(x+7)[/tex]
[tex][x^4(x+7)]+[2x^2(x+7)]+[x+7][/tex]
Notice that each group has common factor [tex](x+7)[/tex]
Factorise the common factor gives
[tex](x+7)[x^4+2x^2+1][/tex]
[tex](x+7)(x^2+1)(x^2+1)[/tex]

The zeros are 
[tex]x+7 = 0[/tex] and four of [tex]x^2+1=0[/tex]
The zero [tex]x^2+1=0[/tex] will give an imaginary results

So, f(x) has one real root and four imaginary root

Answer: B
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Question 3

[tex]f(x) = x^4+16x^2[/tex] 
[tex]f(x) = x^2(x^2+16)[/tex]
The zeros are
[tex]x^2=0[/tex] and [tex]x^2+16=0[/tex]
[tex]x_{1,2} = 0[/tex] and [tex]x_{3, 4} = \sqrt{-16} [/tex]

[tex] \sqrt{-16}= \sqrt{16} \sqrt{-1} [/tex] which gives two answers: [tex]4i[/tex] and [tex]-4i[/tex]

So the factorized form is 
[tex]x^2(x+4i)(x-4i)[/tex]

Answer: C
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Question 4

If [tex]i[/tex] is one of the zeros of [tex]f(x)[/tex] then the other zero would be to conjugate pair of [tex]i[/tex] which is [tex]-i[/tex]

Answer: B
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Question 5

[tex]f(x) = \frac{7}{14-x} [/tex]
The denominator must not equal zero because we can't divide a quantity by zero (undefined result)
So, we need to find the value of [tex]x[/tex] that would make the denominator equal to zero

[tex]14 - x = 0[/tex]
[tex]x = 14[/tex]

Answer: A
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Question 6

[tex]f(x) = \frac{(x-7)(x+4)}{(x^2-4)} [/tex]
To find vertical asymptote, set the denominator to zero then find 'x'

[tex]x^2-4=0[/tex]
[tex]x^2 = 4[/tex]
[tex]x = 2[/tex] and [tex]x = -2[/tex]

Answer: A
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Question 7

[tex]f(x) = \frac{x+9}{x^2+8x+5} [/tex]
The degree of the denominator is higher than the numerator's, hence the horizontal asymptote is y = 0
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Question 8

[tex]f(x) = \frac{x^2+9x-9}{x-9} [/tex]
The degree of numerator is higher than the degree of the denominator, hence no horizontal asymptote

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Question 9

To have a horizontal asymptote, [tex]y = \frac{2}{9} [/tex], both numerator and denominator must be on the same polynomial degree, and the leading degree have coefficient of 2 and 9

Example: [tex] \frac{2x^2+3x+6}{9x^2+6x+9} [/tex]