1.) Let's find out the slope. Checking the Graph, locating two coordinate points. Looking at the graph picking (-4,0) and (0,1)
Inserting those points into that slope formula:
2) According to the second picture. Let's pick two other points. (1,0) and (1,1)
Inserting those points into the formula:
a)0. False.
Thus it can't be 0, since it's not a constant function, and its line it's not parallel to axis x.
b) 1. False 1≠ 1/0
c) True. This quotient is not defined for the set of Real Numbers.
d) False.
3) In a perpendicular line its slope is given by - of another line.
In addition to this we need to know the line general equation
.
From this equation y - 2 = 7/3(x + 5) let's get the slope of our perpendicular line: m= -3/7x
The point (-4, 9) will supply the other two points, let's insert them all
y -9=-3/7(x+4)
C)
4)
a) True. Easily checked by looking at the graph interception.
b) False.
c) False
d) False
5)
B) True. Since the y- intercept of y=|x| is 3.
1.) Let's find out the slope. Checking the Graph, locating two coordinate points. Looking at the graph picking (-4,0) and (0,1)
Inserting those points into that slope formula:
2) According to the second picture. Let's pick two other points. (1,0) and (1,1)
Inserting those points into the formula:
a)0. False.
Thus it can't be 0, since it's not a constant function, and its line it's not parallel to axis x.
b) 1. False 1≠ 1/0
c) True. This quotient is not defined for the set of Real Numbers.
d) False.
3) In a perpendicular line its slope is given by - of another line.
In addition to this we need to know the line general equation
.
From this equation y - 2 = 7/3(x + 5) let's get the slope of our perpendicular line: m= -3/7x
The point (-4, 9) will supply the other two points, let's insert them all
y -9=-3/7(x+4)
C)
4)
a) True. Easily checked by looking at the graph interception.
b) False.
c) False
d) False
5)
B) True. Since the y- intercept of y=|x| is 3.
1. f(x) = 1/4x − 1. B)The line will be less steep. The slope is lower now
2. C)x-intercept (−1/3 , 0). Replacing f(x) = 0 we get x = -1/3
3. B)horizontal line with a slope of zero. The slope is 0 because there is no x in the equation.
4. D)Line changes from decreasing to increasing. The slope changed from negative to positive.
5. See first graph attached
6. See second graph attached. g(x) = (1/3)*x
7. C)The rate of change is different, but the y-intercepts are the same. The rate of change of Mike is negative because he withdraws $25 each week. The y-intercepts of Mike is $200 because that was his initial deposit.
8. B)shifts the line 7 units down. f(x-7) translates f(x) 7 units to the right, given the slope is positive ( = 1) then f(x) is translated 7 units down.
9. No graph is shown
10. C)g(x) = x - 3. g(x) = f(x) - 3, so g(x) is f(x) translated 3 units down.
1. f(x) = 1/4x − 1. B)The line will be less steep. The slope is lower now
2. C)x-intercept (−1/3 , 0). Replacing f(x) = 0 we get x = -1/3
3. B)horizontal line with a slope of zero. The slope is 0 because there is no x in the equation.
4. D)Line changes from decreasing to increasing. The slope changed from negative to positive.
5. See first graph attached
6. See second graph attached. g(x) = (1/3)*x
7. C)The rate of change is different, but the y-intercepts are the same. The rate of change of Mike is negative because he withdraws $25 each week. The y-intercepts of Mike is $200 because that was his initial deposit.
8. B)shifts the line 7 units down. f(x-7) translates f(x) 7 units to the right, given the slope is positive ( = 1) then f(x) is translated 7 units down.
9. No graph is shown
10. C)g(x) = x - 3. g(x) = f(x) - 3, so g(x) is f(x) translated 3 units down.
1) A
2) C
3) B
4) A
5) D
6) C
1) A
2) C
3) B
4) A
5) D
6) C
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The answer is in the image
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