How would I find the range btw im only allowed a Ti-84 calculator So pls tell me how to solve this without using any stepbystep
caulculators or desmos
2 answers:
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Answer:
A. Please see the attached graphs
B. 5. The equation of the graph is y = 3·x - 3
6. The equation of the graph is y = x + 3
7. The equation of the graph is y = 4/5·x - 4
8. The equation of the graph is y = 2·x - 4
9. The graph is the line with equation y = 5·x - 31
10. The graph is the line with equation y = 5·x - 14
11. The graph is the line with equation y = -2·x + 9
12. The graph is the line with equation y = 3·x + 6
Step-by-step explanation:
A. Please see the attached graphs
B. 5. The intercepts are;
(0, -3) and (1, 0)
The slope is given by the following equation;
For the coordinate points, (0, -3) and (1, 0) we have;
m = (0 - (-3))/(1 - 0) = 3
The slope = 3
From the point and slope form, of a straight line equation, we have;
y - 0 = 3(x - 1)
The equation of the graph is therefore;
y = 3·x - 3
The y-intercept occurs at (0, -3)
The x intercept occurs where y = 0
0 = 3·x - 3
x = 3/3 = 1
The x-intercept occurs at (1, 0)
The graph of the equation, y = 3·x - 3, passes through the y and x intercepts (0, -3) and (1, 0) respectively
6. The coordinate points are;
(-3, 0) and (0, -3)
The slope is given by the following equation;
For the coordinate points, (-3, 0) and (0, -3) we have;
m = ((-3) - 0)/(0 - (-3)) = -1
The slope = 1
From the point and slope form, of a straight line equation, we have;
y - 0 = 1(x - (-3)) = x + 3
y = x + 3
The y-intercept occurs at (0, 3)
The x intercept occurs where y = 0
0 = x + 3
x = -3
The x-intercept occurs at (-3, 0)
The graph of the equation, y = x + 3, passes through the y and x intercepts (0, 3) and (-3, 0) respectively
7. The coordinate points are;
(5, 0) and (0, -4)
The slope is given by the following equation;
For the coordinate points, (5, 0) and (0, -4) we have;
m = ((-4) - 0)/(0 - 5) = 4/5
The slope = 4/5
From the point and slope form, of a straight line equation, we have;
y - 0 = 4/5×(x - 5) = -4/5·x - 4
y = 4/5·x - 4
The y-intercept occurs at (0, -4)
The x intercept occurs where y = 0
0 = -4/5·x + 4
-4 = -4/5·x
x = 5
The x-intercept occurs at (5, 0)
The graph of the equation, y = 4/5·x - 4, passes through the y and x intercept (0, 4) and (5, 0) respectively
8. The coordinate points are;
(2, 0) and (0, -4)
The slope is given by the following equation;
For the coordinate points, (2, 0) and (0, -4) we have;
m = ((-4) - 0)/(0 - 2) = 2
The slope =2
From the point and slope form, of a straight line equation, we have;
y - 0 = 2×(x - 2) = 2·x - 4
y = 2·x - 4
The y-intercept occurs at (0, -4)
The x intercept occurs where y = 0
0 = 2·x - 4
4 = 2·x
x = 2
The x-intercept occurs at (2, 0)
The graph of the equation, y = 2·x - 4 passes through the y and x intercept (0, -4) and (2, 0) respectively
C. Using the point and slope form
9. The slope m = 5 and the graph passes through the points (6, -1)
We have the point and slope form given as follows;
y - (-1) = 5·(x - 6)
y = 5·x - 30 - 1 = 5·x - 31
y = 5·x - 31
The graph is the line with equation y = 5·x - 31
10. The slope m = 5 and the graph passes through the points (2, -4)
We have the point and slope form given as follows;
y - (-4) = 5·(x - 2)
y = 5·x - 10 - 4 = 5·x - 14
y = 5·x - 14
The graph is the line with equation y = 5·x - 14
11. The slope m = -2 and the graph passes through the points (4, 1)
We have the point and slope form given as follows;
y - 1 = (-2)·(x - 4)
y = -2·x + 8 + 1 = -2·x + 9
y = -2·x + 9
The graph is the line with equation y = -2·x + 9
12. The slope m = 3 and the graph passes through the points (-3, -3)
We have the point and slope form given as follows;
y - (-3) = 3·(x - (-3))
y = 3·x + 9 - 3 = 3·x + 6
y = 3·x + 6
The graph is the line with equation y = 3·x + 6
