Download Final Exam Review Solutions

Grima MAT 151
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)
let m = 2/3 x1 = 6 y1 = 5
use y – y1 = m(x – x1)
2
𝑦 − 5 = 3 (𝑥 − 6)
2
2
Hint 3 ∗ −6 = 3 ∗
2
y – 5 = 𝑥- 4
3
+5
+5
𝟐
𝟑
Answer #1: 𝒚 = 𝒙 + 𝟏
2) Passing through the points (3,1) and (4,5)
First find the slope
5−1
4
𝑚 = 4−3 = 1 = 4
Next
let m = 4 x1 = 3 y1 = 1
use y – y1 = m(x – x1)
y – 1 = 4(x – 3)
y – 1 = 4x – 12
+1
+1
Answer #2: y = 4x - 11
−6
1
=
−12
3
= −4
#3-4: Find the x and y-intercepts.
3) 𝑦 =
𝑥 2 −5𝑥−6
𝑥+1
Find x-intercept
(You can just set the numerator equal to 0, or you may cross multiply. Both methods give the
same x-intercepts.)
0
1
=
𝑥 2 −5𝑥−6
𝑥+1
Cross multiply
1(x2-5x-6) = 0(x +1)
x2 – 5x – 6 = 0
(x – 6)(x + 1) = 0
x–6=0
x+1=0
x=6
x = -1
y-intercept, let x = 0
𝑦=
02 −5(0)−6
0+1
=
−6
1
= −6
Answer #3: x-intercepts (-1,0) and (6,0) y-intercept (0,-6)
4) x2 + y2 = 121
let y = 0 to find x-intercept
x2 + 02 = 121
x2 = 121
x2 – 121 = 0
(x + 11)(x-11) = 0
x + 11 = 0
x – 11 = 0
x = -11
x = 11
let x = 0 to find y-intercept
02 + y2 = 121
y2 = 121
y2 – 121 = 0
(y + 11)(y-11) = 0
y + 11 = 0
y – 11 = 0
y = -11
y = 11
Answer# 4: x-intercepts (11,0) and (-11,0) y-intercepts (0,11) and (0, -11)
#5-8: find the domain and range.
5)
6)
Domain will be the x-coordinates of each point,
the Range will be the y-coordinates.
Answer
Domain = { 3,4,5,7}
Range = {-1, 2, 6}
Domain [x-coordinate far left point, x-coord far
right point]
Range [y-coordinate of bottom point, ycoordinate of top point]
Answer
Domain [0,5]
Range [-5, 4]
7)
Graph needs to be extended to the left. The
graph goes to the far left edge of the x-axis
which has an x-coordinate of −∞
The domain starts at the far left edge of the xaxis or x = −∞
The domain ends at the point (0,2) The domain
ends at x = 0
Domain (−∞, 0]
Graph extends to the bottom of the y-axis. The
range starts at the bottom of the graph which is
at y = −∞
The range ends at the point (0,2) or more
specifically at y = 0
Range (−∞, 2]
Answer: Domain (−∞, 𝟎]
Range (−∞, 𝟐]
8)
The graph needs to be extended in both
directions. It will be extended to the far left of
the and far right of the x-axis and the bottom and
top of the y-axis.
Answer: do
main (−∞, ∞) range (−∞, ∞)
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
To find the domain, ignore the numerator
Solve denominator = 0
x2 + 6x - 7 = 0
(x + 7)(x – 1) = 0
x+7=0
x = -7
x–1=0
x=1
plot x = −∞, −7,1, ∞ on a number line and create intervals with all round parenthesis.
(−∞, −7)
−∞
−7
(−7,1)
(1, ∞)
1
∞
Answer #9: domain (−∞, −𝟕) ∪ (−𝟕, 𝟏) ∪ (𝟏, ∞)
10) 𝑓(𝑥) = √𝑥 + 5
𝑥+5≥0
-5 - 5
𝑥 ≥ −5 This is the answer, but it is not in interval notation.
Answer #10: [−𝟓, ∞)
11) f(x) = 2x – 6
This is a polynomial and there is no algebra required to find the domain.
Answer #11: 𝒅𝒐𝒎𝒂𝒊𝒏 (−∞, ∞)
12)
12) Find the following:
a) the interval(s) where the function graphed is increasing
The graph increases from the beginning which is x = -∞
To the first turning point x = -3
It also increases from the second turning point
x = -1 to x = ∞
Answer 12a) increasing (−∞, −𝟑) ∪ (−𝟏, ∞)
b) the interval(s) where the function graphed is decreasing
The graph is decreasing between the turning points x = -3
to x = -1
Answer 12b) decreasing (-3,-1)
c) The values of x (if any) where the function has a local
maximum
The x-coord of the high point
12c) answer x = -3
d) The local maximum value (if any)
Asking for y coord of max
12d) answer y = -8
e) The values of x (if any) where the function has a local
minimum
asing for x-coord of low point.
Answer 12e) x = -1
f) The local minimum values (if any)
Wants y-coord of low point.
Answer 12f) y = -12
13) Find the average rate of change of f(x) = x3 + 6x2 from 0 to 2
Create two points. We are given the x-coordinates and need to find y-coordinate.
f(0) = (0)3 + 6(0)2
f(0) = 0 + 0
f(0) = 0
First point (0, 0)
f(2) = (2)3 + 6(2)2
f(2) = 8 + 24
f(2) = 32
Second point (2, 32)
Average rate of change is the slope of the line that connects the points.
32−0
2−0
Average rate of change =
=
32
2
Answer #13: Average rate of change = 16
14) Let 𝑓(𝑥) = 𝑥 2
14a) Find f(x+2) - 4
The x + 2 needs to go inside the parenthesis as it is inside the parenthesis. This will move the graph left
2. The minus 4 will go after the parenthesis and move the graph down.
Answer #14a: f(x + 2) – 4 = (x + 2)2 - 4
14b) Describe how the graph of the given function relates to the graph of 𝑓(𝑥) = 𝑥 2
Answer #14b: left 2 and down 4
15) Let 𝑔(𝑥) = √𝑥
15a) Find -g(x-3) - 5
The negative in front of the g will go in front of the square root. It will reflect the graph over the x-axis.
The x – 3 will go under the radical, and it is okay to put it in a parenthesis, but the parenthesis is not
needed. It will move the graph right.
The minus 5 goes after the square root. It will move the graph down.
Answer #15a: -g(x – 3) – 5 = −√𝒙 − 𝟑 − 𝟓
15b) Describe how the graph of the given function relates to the graph of 𝑔(𝑥) = √𝑥
Answer #15b: reflect over x-axis, right 3 and down 5
16) Let ℎ(𝑥) = |𝑥|
16a) Find h(x-3) + 4
The x – 3 needs to go inside the absolute value, it can be left in a parenthesis but the parenthesis are
not needed. This will move the graph right 3.
The plus 4 will go after the absolute value and move the graph up.
Answer #16a: h(x – 3) + 4 = |𝒙 − 𝟑| + 𝟒
16b) Describe how the graph of the given function relates to the graph of ℎ(𝑥) = |𝑥|
Answer #16b: right 3 and up 4
17) f(x) = x2
17a) Find f(x-3) + 4 and describe the transformation as compared to the function f(x) = x2.
Answer #17a: f(x – 3) + 4 = (x – 3)2 + 4. shifts graph right 3 and up 4.
17b) make a table of values and sketch a graph
x
1
2
3
4
5
f(x – 3) + 4
(1 – 3)2 + 4 = 8
(2 – 3)2 + 4 = 5
(3 – 3)2 + 4 = 4
(4 – 3)2 + 4 = 5
(5 – 3)2 + 4 = 8
point
(1,8)
(2,5)
(3,4)
(4,5)
(5,8)
17c) state the domain and range of the function
Domain – the graph extends from the far left to the far right of the x-axis unbroken so the domain is (−∞, ∞)
Range – the y-coordinate of the bottom of the graph is y = 4. The y-coordinate of the top is ∞ so the range is
[4, ∞)
Answer #17c: domain (−∞, ∞) range [𝟒, ∞)
17d) state the intervals where the function in increasing and decreasing
the graph increases from the x-coordinate of the vertex (x= 3) to the end of the graph ( x = ∞)
The graph decreases from the start of the graph ( x = ∞) to the x-coordinate of the vertex (x = 3)
Answer #17d: increasing (𝟑, ∞) decreasing (−∞, 𝟑)
17e) state if the function has a local maximum, if it does state the local maximum value
Answer #17e: The function has no local minimum
17f) state if the function has a local minimum, if it does state the local minimum value
Answer #17f: There is a local minimum (the vertex) at x = 3, the local minimum value is y = 4.
18a) Use synthetic division to factor the polynomial, f(x) = x3 + x2 + 2x + 2
here is a graph of f(x)
I will use the number (-1) for my synthetic division as it is the only x-intercept.
since x = -1 is a zero, I know (x+1) is a factor of f(x)
the synthetic division will get me the remaining factors.
1
1
-1
1
0
The result of my synthetic division gives me
-1
𝑥 3 +𝑥 2 +2𝑥+2
𝑥+1
2
-2
0
= 𝑥 2 + 2 𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟 0
so now I can factor f(x)
f(x) = x3 + 2x2 – 5x – 6 = (x+1)(x2+2)
Answer #18a: f(x) = (x+1)(x2 + 2)
18b) Solve x3 + x2 + 2x + 2= 0
(x + 1)(x2 + 2) = 0
x+1=0
x = -1
2
0
2
x2 + 2 = 0
x2 = - 2
√𝑥 2 = ±√−2
𝑥 = ±√2𝑖
Answer #18b: x = -1, 𝒙 = ±√𝟐𝒊
(the x2 + 2 is prime, so this is completely factored)
#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
19a) Domain
x- 3 = 0
x=3
Answer: all real numbers except 3
19b) Vertical Asymptote (if any)
Answer: x = 3
19c) Horizontal asymptote, or slant asymptote
2
𝑦= =2
1
Answer: y = 2
19d) x- intercept(s) if any
2x + 6 = 0
2x = -6
x = -3
Answer (-3,0)
19e) y-intercept(s) if any
𝑓(0) =
2∗0+6
0−3
6
= −3 = −2
Answer #19e: (0,-2)
19f) Sketch a graph of the function : show all the features found in parts b – e
y
8
6
4
2
x
-18
-16
-14
-12
-10
-8
-6
-4
-2
2
(-3,0)
-2
(0,-2)
-4
-6
-8
4
6
8
10
12
14
16
18
20. f ( x) 
x 1
x  4 x  21
2
20a) Domain
x2 + 4x – 21 = 0
(x+7)(x-3) = 0
x +7=0 x–3 =0
x = -7 x = 3
Answer: all real numbers except -7, 3
20b) Vertical Asymptote (if any)
Answer: x = -7 and x = 3
20c) Horizontal asymptote, or slant asymptote
There is no work to find the horizontal
asymptote. The largest exponent is in the
denominator so x-axis which has the equation y =
0 is the horizontal asymptote.
Answer: y = 0 (the x-axis)
20d) x- intercept(s) if any
It’s enough to just set the numerator equal to
zero.
x–1=0
x=1
Answer: (1,0)
20e) y-intercept(s) if any
0−1
−1
1
𝑓(0) = 02 +4∗0−21 = −21 = 21
Answer #20e: (0, 1/21)
20f) Sketch a graph of the function : show all the features found in parts b - e
#21-22: Write each side with the same base then solve. Be sure to check your answer.
21) 3x-3 = 27
3x-3 = 27
3x-3 = 33
x–3=3
Answer #21: x = 6
1 𝑥−2
2
22) ( )
1 𝑥−2
(2)
1 𝑥−2
(2)
=
1
8
13
= 23
1 3
= (2)
x– 2 = 3
Answer #22: x = 5
#23-26: Solve the logarithmic equations, round to 2 decimals when needed.
23) log2(x+1) = 5
log2(x+1) = 5
25 = x + 1
32 = x + 1
Answer #23: x = 31
24) log4(x-5)=3
Log4(x-5) = 3
43 = x – 5
64 = x – 5
Answer #24: x = 69
25) log2 (x+14) – log2 (x+6)= 1
𝑙𝑜𝑔2
21 =
2
1
=
𝑥+14
𝑥+6
=1
𝑥+14
𝑥+6
𝑥+14
𝑥+6
2(x +6) = 1(x+14)
2x + 12 = x + 14
2x – x = 14 – 12
Answer #25: x = 2
26) log2 (x+2)+log2 (x+6) = 5
Log2(x+2)(x+6) = 5
Log2(x2 + 8x+ 12) = 5
25 = x2 + 8x+ 12
32 = x2 + 8x + 12
0 = x2 + 8x – 20
0 = (x + 10)(x – 2)
x+ 10 = 0
x = -10 (doesn’t check)
Answer #26: x = 2
x–2=0
x=2
Final Exam Review
27) Solve using the substitution method
4
4
1
x y 5
5
4
x  y 1
1
20 ∗ 5 𝑥 + 20 ∗ 4 𝑦 = 20 ∗ 5
16x + 5y = 100
16(y+1) + 5y = 100
16y + 16 + 5y = 100
21y + 16 = 100
21y = 84
y = 84/21
y=4
Solve for x
x=y+1
x=4+1
x=5
Answer #27: (5,4)
2
28) Solve using the elimination method.
4(2x + 3y = 13)
3(5x – 4y = 21)
8x + 12y = 52
15x – 12y = 63
23x = 115
x = 115/23
x=5
Solve for y:
2x + 3y = 13
2(5) + 3y = 13
10 + 3y = 13
3y = 3
y=1
Answer #28: (5,1)
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)
Step 1:
2(−𝑥 + 𝑦 + 2𝑧 = 7)
2𝑥 + 4𝑦 − 5𝑧 = −5
-2x + 2y + 4z = 14
2x + 4y – 5z = -5
6y - z = 9
Step 2:
6y – z = 9
3(-2y + 5z = 11)
6y –z = 9
-6y + 15z = 33
14z = 42
z=3
Solve for y
6y – 3 = 9
6y = 12
y=2
Step 3
Solve for x
𝑥 − 3𝑦 + 3𝑧 = 4
x – 3(2) + 3(3) = 4
x–6+9=4
x+3=4
x=1
Answer #29: (3,2,1)
−𝑥 + 𝑦 + 2𝑧 = 7
𝑥 − 3𝑦 + 3𝑧 = 4
-2y + 5z = 11
#30-31: Solve the following systems of equations.
30)
x  7 y  4
x 2  y 2  10
Solve top for x
x – 7y = -4
+7y + 7y
x = 7y – 4
substitute into bottom
(7y-4)2 + y2 = 10
(7y -4)(7y – 4) + 1y2 = 10
49y2 – 28y – 28y + 16 + 1y2 = 10
50y2 – 56y + 16 = 10
50y2 – 56y + 6 = 0
Divide by 2
25y2 – 28y + 3 = 0
Factor
(25y – 3)(y – 1) = 0
25y – 3 = 0
25y = 3
y = 3/25
y–1=0
y=1
Solve for x
x = 7y – 4
x = 7(3/25) – 4
used calculator
x = -79/25
x = 7(1) – 4
x=3
Answer #30: (-79/25, 3/25) (3,1)
31)
x  2 y  11
x 2  y  13
Solve bottom equation for y
x2 + y = 13
-x2
- x2
y = 13 – x2
Substitute into top equation
x + 2(13 – x2) = 11
x + 26 – 2x2 =
11
-x - 26 +2x2 +2x2 - x -26
0 = 2x2 – x – 15
0 = (2x + 5)(x - 3)
2x + 5 = 0
x–3= 0
2x = -5
x=3
x = -5/2
Solve for y
y = 13 – x2
y = 12 – (-5/2)2
solved on my calculator
y = 13 – 32
y=4
y = 27/4
Answer #31: (-5/2, 27/4) (3,4)
32) Find the difference quotient for the function f(x) = x2 +3x - 5:
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
f(x+h) = (x + h)2 + 3(x+h) – 5
f(x + h) = (x + h)(x + h) + 3(x + h) – 5
f(x + h) = x2 + 1xh + 1xh + h2 + 3x + 3h - 5
f(x + h) = x2 + 2xh + h2 + 3x + 3h -5
(𝑥 2 +2𝑥ℎ+ℎ 2 +3𝑥+3ℎ−5)−(𝑥 2 +3𝑥−5)
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
=
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
=
𝑥 2 +2𝑥ℎ+ℎ 2 +3𝑥+3ℎ−5−𝑥 2 −3𝑥+5
ℎ
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
=
2𝑥ℎ+ℎ 2 +3ℎ
5
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
=
ℎ(2𝑥+ℎ+3)
ℎ
Answer #32:
ℎ
𝒇(𝒙+𝒉)−𝒇(𝒙)
𝒉
= 𝟐𝒙 + 𝒉 + 𝟑
Answers:
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
14a) f(x+2) – 4 = (x + 2)2 - 4
14b) Same shape as f(x) = x2, except moved down 4 units and left 2 units
15a) -g(x-3) – 5 = −√𝑥 − 3 − 5
15b) Same shape as g(x) = √𝑥, except moved down 5 units, right 3 units and reflected over x-axis
16a) h(x-3) + 4 = |𝑥 − 3| + 4
16b) Same shape as h(x) = |𝑥|, except shifted up 4 units and right 3 units
17a) f(x-3) + 4 = (x-3)2 + 4
The graph has the same shape as f(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
18a)
f(x) = (x+1)(x2 + 2)
18b) x = -1, ±𝑖√2
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
2
3
4
5
6
7
8
9
10
11
12
13
14
-1
-2
-3
-4
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)