Answer Key - David Beydler`s Math

Algebra Review – Homework Answer Key
Math 140 – Prof. Beydler
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Homework Answer Key
2
1. Find an equation of the line through (0, βˆ’4) with slope 3. Then rewrite the equation in slopeintercept form. Finally, graph the line.
𝟐
π’š + πŸ’ = πŸ‘ (𝒙 βˆ’ 𝟎) (This is point-slope form.)
𝟐
π’š = πŸ‘ 𝒙 βˆ’ πŸ’ (This is slope-intercept form.)
2. Find an equation of the line through (2, 5) with slope βˆ’1. Then rewrite the equation in slopeintercept form. Finally, graph the line.
π’š βˆ’ πŸ“ = βˆ’πŸ(𝒙 βˆ’ 𝟐) (This is point-slope form.)
π’š = βˆ’π’™ + πŸ• (This is slope-intercept form.)
1
2
3. Find an equation of the line through (βˆ’1, 2) and (0 , βˆ’ 3). Then rewrite the equation in slopeintercept form.
𝟏
πŸ•
π’š βˆ’ 𝟐 = βˆ’ πŸ” (𝒙 + 𝟏)
πŸ•
𝟐
π’š = βˆ’πŸ”π’™ βˆ’ πŸ‘
𝟐
πŸ•
(This is point-slope form. You could also write π’š + πŸ‘ = βˆ’ πŸ” (𝒙 βˆ’ 𝟎).)
(This is slope-intercept form.)
4. Graph the line π‘₯ = βˆ’2. What is the slope of the line?
The slope is undefined.
Math 140
Algebra Review – Homework Answer Key
5. Graph the line 𝑦 = 6. What is the slope of the line?
The slope is 0.
6. Write an equation for the following line. What is the slope of the line?
Equation: π’š = 𝟏
Slope: 𝟎
7. Graph 𝑓 (π‘₯ ) = π‘₯ 2 βˆ’ 2 by finding and plotting the vertex, π‘₯-intercepts, and 𝑦-intercept.
Vertex: (𝟎, βˆ’πŸ)
𝒙-intercepts: (√𝟐, 𝟎) and (βˆ’βˆšπŸ, 𝟎)
π’š-intercept: (𝟎, βˆ’πŸ)
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Algebra Review – Homework Answer Key
Math 140
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8. Graph 𝑓 (π‘₯ ) = 2π‘₯ 2 + 4π‘₯ βˆ’ 6 by finding and plotting the vertex, π‘₯-intercepts, and 𝑦-intercept.
Vertex: (βˆ’πŸ, βˆ’πŸ–)
𝒙-intercepts: (βˆ’πŸ‘, 𝟎) and (𝟏, 𝟎)
π’š-intercept: (𝟎, βˆ’πŸ”)
1 βˆ’ 3π‘₯ if π‘₯ ≀ 4
2
. Find 𝑓(8), 𝑓(0), and 𝑓(βˆ’1).
if π‘₯ > 4
π‘₯
𝒇(𝟎) = 𝟏 𝒇(βˆ’πŸ) = πŸ’
9. Suppose 𝑓 (π‘₯ ) = {
𝟏
𝒇 (πŸ–) = πŸ’
√π‘₯ + 14 if π‘₯ < βˆ’6
10. Suppose 𝑓 (π‘₯ ) = { 2π‘₯ 2 + 1 if βˆ’ 6 ≀ π‘₯ ≀ βˆ’1. Find 𝑓(βˆ’10), 𝑓(βˆ’6), 𝑓 (βˆ’1), 𝑓 (0), and 𝑓 (27).
3
if π‘₯ > βˆ’1
√π‘₯
𝒇(βˆ’πŸπŸŽ) = 𝟐
𝒇(βˆ’πŸ”) = πŸ•πŸ‘
𝒇(βˆ’πŸ) = πŸ‘
𝒇 (𝟎) = 𝟎
𝒇(πŸπŸ•) = πŸ‘
11. Find 𝑓(π‘₯ + 1) where 𝑓 (π‘₯ ) = π‘₯ 2 βˆ’ 3π‘₯ + 2. Simplify.
𝒇 (𝒙 + 𝟏) = 𝒙 𝟐 βˆ’ 𝒙
π‘₯
12. Find 𝑓(π‘₯ + β„Ž) where 𝑓 (π‘₯ ) = π‘₯+1.
𝒙+𝒉
𝒇(𝒙 + 𝒉) = 𝒙+𝒉+𝟏
13. Find
𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)
β„Ž
𝒇(𝒙+𝒉)βˆ’π’‡(𝒙)
𝒉
14. Find
=𝟐
𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)
β„Ž
𝒇(𝒙+𝒉)βˆ’π’‡(𝒙)
𝒉
where 𝑓 (π‘₯ ) = 2π‘₯ βˆ’ 3. Simplify.
where 𝑓 (π‘₯ ) = π‘₯ 2 . Simplify.
= πŸπ’™ + 𝒉
π‘₯+1
15. Find the domain of 𝑓 (π‘₯ ) = π‘₯2 βˆ’π‘₯βˆ’2.
All real numbers except -1 and 2.
16. Find the domain of 𝑓 (π‘₯ ) = √6 βˆ’ 2π‘₯.
All real numbers such that 𝒙 ≀ πŸ‘.
17. Find the domain of 𝑓 (π‘₯ ) =
π‘₯βˆ’1
√π‘₯+2
.
All real numbers such that 𝒙 > βˆ’πŸ.
Algebra Review – Homework Answer Key
Math 140
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18. Factor completely.
a) 3π‘₯ 4 + 4π‘₯ 3
π’™πŸ‘ (πŸ‘π’™ + πŸ’)
b) 12π‘₯ 2 + 36π‘₯ βˆ’ 48
𝟏𝟐(𝒙 βˆ’ 𝟏)(𝒙 + πŸ’)
c) 6π‘₯ 2 + 6π‘₯ βˆ’ 12
πŸ”(𝒙 βˆ’ 𝟏)(𝒙 + 𝟐)
d) π‘₯ 2 βˆ’ 2π‘₯ + 1
(𝒙 βˆ’ 𝟏)𝟐
e) 4π‘₯ 2 βˆ’ 9
(πŸπ’™ βˆ’ πŸ‘)(πŸπ’™ + πŸ‘)
f) π‘₯ 3 βˆ’ 4π‘₯ 2 βˆ’ 3π‘₯ + 12
(𝒙 βˆ’ πŸ’)(π’™πŸ βˆ’ πŸ‘)
19. Rewrite the following using rational exponents.
𝟏
a) √π‘₯ = π’™πŸ
5
e) π‘₯2 = πŸ“π’™
βˆ’πŸ
𝟏
4
b) √π‘₯ = π’™πŸ’
1
f) π‘₯5 = 𝒙
βˆ’πŸ“
g)
1
√π‘₯
20. Rewrite the following using radicals or fractions.
1
πŸ‘
a) π‘₯ 3 = βˆšπ’™
𝟏
e) π‘₯ βˆ’3 = π’™πŸ‘
3
πŸ’
b) π‘₯ 4 = βˆšπ’™πŸ‘
2
f) π‘₯ βˆ’3 =
𝟐
5
c) √π‘₯ 2 = π’™πŸ“
=𝒙
𝟏
𝟐
3
d) = πŸ‘π’™βˆ’πŸ
π‘₯
βˆ’
1
c) π‘₯ 2 = βˆšπ’™
𝟏
πŸ’
d) 4π‘₯ βˆ’1 = 𝒙
πŸ‘
βˆšπ’™ 𝟐
21. Suppose the demand function for π‘₯ hundred units of a textbook is:
𝐷(π‘₯ ) = βˆ’0.37π‘₯ + 47 (in dollars)
and the cost of producing π‘₯ hundred textbooks is:
𝐢 (π‘₯ ) = 1.38π‘₯ 2 + 15.15π‘₯ + 115.5 (in hundreds of dollars).
Find the revenue 𝑅(π‘₯ ) and profit 𝑃 (π‘₯ ).
𝑹(𝒙) = βˆ’πŸŽ. πŸ‘πŸ•π’™πŸ + πŸ’πŸ•π’™
𝑷(𝒙) = βˆ’πŸ. πŸ•πŸ“π’™πŸ + πŸ‘πŸ. πŸ–πŸ“π’™ βˆ’ πŸπŸπŸ“. πŸ“
22. Suppose that when the price of a water bottle is 𝑝 dollars per unit, then π‘₯ thousand units will be
purchased by consumers, where 𝑝 = βˆ’0.01π‘₯ + 4. The cost of producing π‘₯ thousand bottles is
𝐢 (π‘₯ ) = 0.005π‘₯ 2 + π‘₯ + 120 thousand dollars.
a) Find the profit function, 𝑃(π‘₯). (Notice that big 𝑃 and little 𝑝 mean two different things. Like
passwords, math is case sensitive! )
𝑷(𝒙) = βˆ’πŸŽ. πŸŽπŸπŸ“π’™πŸ + πŸ‘π’™ βˆ’ 𝟏𝟐𝟎
b) Using 𝑃(π‘₯), determine the level of production π‘₯ that results in maximum profit.
𝒙 = 𝟏𝟎𝟎 thousand bottles = 𝟏𝟎𝟎, 𝟎𝟎𝟎 bottles
Math 140
Algebra Review – Homework Answer Key
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c) What unit price 𝑝 corresponds to maximum profit?
𝒑 = $πŸ‘ per bottle
23. You start a company that produces mechanical pencils whose lead does not break while writing.
Your total cost consists of a fixed overhead of $6000 plus production costs of $5 per pencil.
a) Find the total cost function 𝐢(π‘₯), where π‘₯ is the number of pencils produced.
π‘ͺ(𝒙) = πŸ”πŸŽπŸŽπŸŽ + πŸ“π’™
b) Sketch the graph of 𝐢(π‘₯).
24. You’re in the cut-throat iPhone case business. You estimate that you can sell a case for $2 more
than it costs to produce it. Your fixed costs are $11500.
a) Express total profit 𝑃(π‘₯) as a function of the level of production π‘₯ (that is, π‘₯ represents the
number of cases produced).
𝑷(𝒙) = πŸπ’™ βˆ’ πŸπŸπŸ“πŸŽπŸŽ
b) How much profit (or loss) is generated when 5000 cases are produced?
𝑷(πŸ“πŸŽπŸŽπŸŽ) = βˆ’$πŸπŸ“πŸŽπŸŽ (loss)
c) What is the level of production at the break-even point? (Hint: Recall that the break-even
point occurs when 𝑅(π‘₯ ) = 𝐢 (π‘₯ ). Equivalently, it happens when 𝑃(π‘₯ ) = 0.)
𝒙 = πŸ“πŸ•πŸ“πŸŽ cases
25. Suppose the supply and demand functions are 𝑆(π‘₯ ) = 4π‘₯ + 200 and 𝐷 (π‘₯ ) = βˆ’3π‘₯ + 480,
respectively. Find the value of π‘₯𝑒 for which equilibrium occurs and the corresponding equilibrium
price 𝑝𝑒 .
𝒙𝒆 = πŸ’πŸŽ and 𝒑𝒆 = πŸ‘πŸ”πŸŽ