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Math 1031 College Algebra and Probability
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Midterm 2 Review
Chapter 2 and 3.1-3.4
2.1: Coordinate Geometry
• Memorize 2 formulas for two points (x1 , y1 ) and (x2 , y2 ):
Midpoint formula:
x1 + x2 y1 + y2
,
2
2
p
(x2 − x1 )2 + (y2 − y1 )2
Distance formula:
d=
1. Find the midpoint and distance between (−3, 4) and (6, −2).
Solution: Midpoint:
3
2, 1
√
, Distance: 3 13
2. Find the midpoint and distance between
Solution: Midpoint:
13 −9
4 , 2
7
2 , −5
and (3, −4).
√
, Distance:
5
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2.2: Linear Equations and Inequalities
• Be able to find x and y intercepts and be able to describe in words what x and y intercepts are.
3. Graph 6x + 8y ≤ −12.
Solution: x intercept at (−2, 0) and y intercept at (0, −3/2). Connect the intercepts using a solid
line. Shade the half-plane that does not include the origin (0, 0).
4. Graph 4x − 3y < 16.
Solution: x intercept at (4, 0) and y intercept at 0, −16
. Connect the intercepts using a dotted line.
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Shade the half-plane that include the origin (0, 0).
2.3: Determining the Equation of a Line
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• Memorize characteristics of 3 types of equations of lines:
– Point-Slope Form: (y − y1 ) = m(x − x1 )
◦ m is the slope
◦ x and y are the most general x and y
◦ (x1 , y1 ) is any specific point on the line
– Slope-Intercept Form: y = mx + b
◦ m is the slope
◦ x and y are the most general x and y
◦ b is the y-intercept
– Standard Form: Ax + By = C
◦ A, B, and C are all integers, A ≥ 0
◦ x and y are the most general x and y
• Memorize Property 2.1
– Two lines with slopes m1 and m2 are:
◦ parallel if m1 = m2
−1
◦ perpendicular if m1 = m
2
5. Find the equation of the line determined by the points (3, 3) and (5, −7). Put your answer in slopeintercept form.
Solution: y = −5x + 18
6. Find the equation of the line which has y intercept −2 and slope 3/4. Put your answer in standard
form.
Solution: 3x − 4y = 8
7. Find the equation of the line parallel to 4x − 5y = 15 that goes through the point (6, 13). Put your
answer in point-slope form.
Solution: (y − 13) = 45 (x − 6)
8. Find the equation of the line perpendicular to y = − 85 x − 1 through the point (2, 1). Put your answer
in standard form.
Solution: 8x − 5y = 11
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2.4: Graphing Techniques
• Memorize rules for determining symmetry
– A graph has y-axis symmetry if replacing x with -x results in an equivalent equation
– A graph has x-axis symmetry if replacing y with -y results in an equivalent equation
– A graph has origin symmetry if replacing both x with -x and y with -y results in an equivalent
equation
• Be familiar with common restrictions:
– Avoid
√
anything
and negative
0
• Look at resulting values in common graphs:
√
– ◦ positive ≥ 0
– ◦ (anything)2 ≥ 0
– ◦ |anything| ≥ 0
9. Graph y = −2|x − 3| + 1.
Solution:
• No symmetry
• x intercepts at ( 52 , 0) and ( 27 , 0), y intercept at (0, −5)
• No restrictions on x. y ≤ 1
• The graph of y = |x| is shifted reflected across the x-axis and becomes twice as narrow. Then it
is shifted 1 unit up and 3 units right.
10. Graph y =
√
x2 + x − 2.
Solution:
• No symmetry
• x intercepts at (−2, 0) and (1, 0), no y intercept
• x ≤ −2 or x ≥ 1 and y ≥ 0
• You have to plot points since there isn’t really a basic graph we’re working with here.
11. Graph y =
x+1
.
x2 + x − 6
Solution:
• No symmetry
• x intercept at (−1, 0), y-intercept at 0, − 16
• Restrictions are x 6= 2 and x 6= −3.
• You have to plot points since there isn’t really a basic graph we’re working with here.
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2.5: Circles
• Memorize the standard form for the equation of a circle with center at (h, k) and radius r.
(x − h)2 + (y − k)2 = r2
• No ellipses or hyperbolas will be on the exam
• Complete the square with two variables
12. Write the equation for this circle in standard form. Give the center and radius then graph the circle.
2x2 − 16x + 2y 2 + 20y + 58 = 0
√
Solution: Center at (4, −5) and radius 2 3
13. Write the equation for this circle in standard form. Give the center and radius then graph the circle.
1 2
1
x − 6x + y 2 + 2y + 36 = 0
4
4
Solution: Center at (12, −4) and radius 4
14. Write the equation for this circle in standard form. Give the center and radius then graph the circle.
x2 − x + y 2 + 3y −
Solution: Center at
1
3
2, −2
√
and radius 2 2
11
=0
2
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3.1: Functions
• Be able to evaluate a function, including piecewise functions.
• Domain is the set of all possible inputs x
• Range is the set of all possible outputs f (x)
• A function is even if it has y-axis symmetry. That is, a function is even if f (−x) = f (x)
• A function is odd if it has origin symmetry. That is, a function is odd if f (−x) = −f (x).
15. If
g(x) =
2x − 1
x2
if x ≥ 4
,
if x < 4
find g(6), g(4), and g(0).
Solution: g(6) = 11, g(4) = 7, g(0) = 0
16. Give the domain and range of the following functions. Is each function odd, even, or neither?
√
◦ f (x) = x − 2
Solution: Domain: {x|x ≥ 2}, Range: {f (x)|f (x) ≥ 0}, neither odd nor even
◦ g(x) = (x − 1)3
Solution: Domain: All real numbers, Range: All real numbers, neither odd nor even
◦ h(x) =
√
x2 + 1 − 4
Solution: Domain: All real numbers, Range: {f (x)|f (x) ≥ −3}, even
3.2: Linear Functions
• Be comfortable graphing a line in slope-intercept form
• Be able to explain what f (x) means and what f (2) means
17. Graph f (x) = − 41 x + 7. What is f (4)?
Solution: The y intercept is at (0, 7) and the x intercept is at (28, 0). f (4) = 6
18. Stephen makes $500 each month plus $50 for every cell phone he sells. Write a linear function that
describes his situation. How much money does he make if he sells 4 cell phones? What is the domain?
Solution: f (x) = 50x + 500. He will make $700 if he sells 4 cell phones. The domain is non-negative
integers (0, 1, 2, 3, ...) since he can’t sell 3.5 or −2 cell phones.
3.3: Quadratic Functions
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• Be able to explain what each part of vertex form y = a(x − h)2 + k means
◦ y = x2 + k is a vertical shift
◦ y = ax2 is a stretch, scrunch, or flip depending on the value of a
◦ y = (x − h)2 is a horizontal shift
• Be able to graph a parabola in vertex form
• Complete the square to write f (x) = ax2 + bx + c in the form f (x) = a(x − h)2 + k.
19. Graph f (x) = −(x − 5)2 + 3
Solution: The basic graph of y = x2 is flipped across the x axis so it now opens downward, then
moved 5 units right and 3 units up.
20. The basic graph of f (x) = x2 is scrunched (made narrower) by a factor of 2, then is moved 6 units
down and 3 units left. Write the new quadratic function this new graph illustrates.
Solution: f (x) = 2(x + 3)2 − 6
21. Put f (x) = 3x2 + 6x − 1 in the form f (x) = a(x − h)2 + k, then graph the function.
Solution: f (x) = 3(x + 1)2 − 4. The parabola is made narrower by a factor of 3, moved 4 units down
and 1 unit right.
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3.4: More Quadratic Functions
• Be able to graph a parabola using the techniques from this chapter
• Memorize the formula for the vertex of the parabola f (x) = ax2 + bx + c
b
b
− , f (− )
2a
2a
• Be able to find the minimum or maximum value of a function based on a story problem.
22. What is the vertex of f (x) = −2x2 − 4x + 2? Does this vertex represent a minimum or a maximum?
Solution: The vertex is at
a<0
−4
, 4(−2)(2)−(−4)
− 2(−2)
4(−2)
2
= (−1, 4) and it represents a maximum since
23. You have 1200 feet of fencing available and you want to make a rectangular pen which is divided into 5
smaller pens of equal size. The fencing used to divide the larger area into smaller pens must be parallel
to 2 sides of the rectangle. Write a quadratic equation that illustrates this situation. What are the
dimensions of the rectangular pen that will maximize the area?
Solution: The total length of fence is 1200 feet. This will build four sides of a rectangle plus four lines
subdividing the rectangle, for a total of two sides of one length and six sides of a second length. This
gives 6a + 2b = 1200. Solving for b, we get b = 21 (1200 − 6a) = 600 − 3a. So the two sides are a and
600−3a. The area of the pen is given by f (a) = a(600−3a) = −3a2 +600a. The vertex is at (100, 30000)
and so the value of a that maximizes the area is a = 100. Then b = 600 − 3a = 600 − 300 = 300, so
the dimensions of the pen that miximize the area are 100 feet by 300 feet.