# final exam review ```Final exam review
#1-2: find an equation for the line with the given properties. Express your answer in slope
intercept form.
2
1) Slope = 3; containing the point (6,5)
2) Passing through the points (3,1) and (4,5)
#3-4: Find the x and y-intercepts.
3) 𝑦 =
𝑥 2 −5𝑥−6
𝑥+1
4) x2 + y2 = 121
#5-8: find the domain and range.
5)
6)
7)
8)
Final exam review
#9-11: Use algebra to find the domain of each function. Write your answer in interval notation
𝑥−4
9) 𝑓(𝑥) = 𝑥 2 +6𝑥−7
10) 𝑓(𝑥) = √𝑥 + 5
11) f(x) = 2x – 6
#12: Find the following:
a) the interval(s) where the function graphed is increasing
b) the interval(s) where the function graphed is decreasing
c) The values of x (if any) where the function has a local maximum
d) The local maximum value (if any)
e) The values of x (if any) where the function has a local minimum
f) The local minimum values (if any)
13) Find the average rate of change of f(x) = x3 + 6x2 from 0 to 2
#14-16: describe how the graph of the given function relates to the graph of a common function
14) f(x) = (x+2)2 - 4
15) f(x) = −√𝑥 − 3 – 5
16) 𝑓(𝑥) = |𝑥 − 3| + 4
Final Exam Review
17) f(x) = (x – 3)2 + 4
a) describe the transformation as compared to the function f(x) = x2, specifically state if the graph is shifted left,
right, up, down and if any reflection has occurred
b) make a table of values and sketch a graph
c) state the domain and range of the function
d) state the intervals where the function in increasing and decreasing
e) state if the function has a local maximum, if it does state the local maximum value
f) state if the function has a local minimum, if it does state the local minimum value
18) Use synthetic division to factor the polynomial, then solve the equation.
x3 + x2 + 2x + 2= 0
#19 – 20: For each problem find the following:
a) Domain
b) Vertical Asymptote (if any)
c) Horizontal asymptote, or slant asymptote
d) x- intercept(s) if any
e) y-intercept(s) if any
f) Sketch a graph of the function : label all the features found in parts b - e
19. f ( x ) 
2x  6
x3
20. f ( x) 
x 1
x  4 x  21
2
#21-22: Write each side with the same base then solve. Be sure to check your answer.
21) 3x-3 = 27
1 𝑥−2
22) (2)
1
=8
#23-26: Solve the logarithmic equations, round to 2 decimals when needed.
23) log2(x+1) = 5
24) log4(x-5)=3
25) log2 (x+14) – log2 (x+6)= 1
26) log2 (x+2)+log2 (x+6) = 5
Final Exam Review
27) Solve using the substitution method
28) Solve using the elimination method.
4
1
x y 5
5
4
x  y 1
2 x  3 y  13
5 x  4 y  21
29) Solve each system of equations, by hand without matrices
2𝑥 + 4𝑦 − 5𝑧 = −5
−𝑥 + 𝑦 + 2𝑧 = 7
𝑥 − 3𝑦 + 3𝑧 = 4
(pair the middle equation with the other 2 and drop out the x’s)
#30-31: Solve the following systems of equations.
30)
x  7 y  4
x  y  10
2
2
31)
x  2 y  11
x 2  y  13
32) Find the difference quotient for the function f(x) = x2 +3x - 5:
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
2
1) 𝑦 = 3 𝑥+1
2) y = 4x – 11
3) x-intercepts (-1,0) and (6,0) y-intercept (0,-6)
4) x-intercepts (11,0) and (-11,0) y-intercepts (0,11) and (0, -11)
5) domain {3,4,5,7} Range {-1,2,6}
7) 𝑑𝑜𝑚𝑎𝑖𝑛 (−∞, 0] 𝑟𝑎𝑛𝑔𝑒 (−∞, 2]
6) Domain [0,5] range [-5,4]
8) 𝑑𝑜𝑚𝑎𝑖𝑛 (−∞, ∞) 𝑟𝑎𝑛𝑔𝑒 (−∞, ∞)
9) domain (−∞, −7) ∪ (−7,1) ∪ (1, ∞)
10) 𝑑𝑜𝑚𝑎𝑖𝑛 [−5, ∞)
11) 𝑑𝑜𝑚𝑎𝑖𝑛 (−∞, ∞)
12a) (−∞, −3) ∪ (−1, ∞)
12b) (-3,-1)
12e) x = -1
13) average rate of change = 16
12f) y = -12
12c) x = -3
12d) y = -8
14) same shape as g(x) = x2, except moved down 4 units and left 2 units
15) same shape as g(x) = √𝑥, except moved down 5 units, right 3 units and reflected over x-axis
16) same shape as g(x) = |𝑥|, except shifted up 4 units and right 3 units
17a) The graph has the same shape as g(x) = x2, except it is shifted right 3 units and up 4 units.
17b)
x
5
4
3
2
1
f(x) or y
8
5
4
5
8
Computation, use calculator to get y - column
(5-3)2 + 4
(4-3)2 + 4
(3-3)2 + 4
(2-3)2 + 4
(1-3)2 + 4
17c) Domain (−∞, ∞) Range [4, ∞) (see me for help if you need some finding the domain and range)
17d) The graph is increasing (3, ∞) and decreasing from (−∞, 3)
17e) The graph does not have a high point so it has no local maximum
17f) The low point is the local minimum. We say there is a local minimum at x = 3 and the local
minimum value is y = 4
x = -1, &plusmn;𝒊√𝟐
18)
19a) all real numbers except 3
19d) (-3,0)
19b) x = 3
19c) y = 2
19e) (0, -2)
19f)
y
8
6
4
2
x
-18
-16
-14
-12
-10
-8
-6
-4
-2
2
(-3,0)
4
6
8
10
12
14
16
18
-2
(0,-2)
-4
-6
-8
20a) all real numbers except -7,3
20b) x = -7 and x = 3
20c) y = 0 (the x-axis) 11d) (1,0)
20e) (0, 1/21)
20f)
y
4
3
2
1
x
-14 -13 -12 -11 -10
-9
-8
-7
-6
-5
-4
-3
-2
-1
1
-1
-2
-3
-4
2
3
4
5
6
7
8
9
10
11
12
13
14
21) x = 6
22) x = 5
23) x = 31
24) x = 69
25) x = 2
26) x = 2
27) (5,4)
28) (5,1)
29) (1,2,3)
30) (-79/25, 3/25) and (3,1)
32) 2x + h+ 3
31) (-5/2, 27/4) and (3,4)
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