Asked by G
My book uses the slope y-intercept method for graphing.
I am trying to graph the following:
y>3/2x-5/2
(sign > is greater then or equal to)
The way I understand it I would start on the Y axis at 0 and go down 5 and then to the right 2 for the -5/2. Then I would go up 3 and to the right 2 for the 3/2. The book shows(0,3/2) and (2,-7/2) as the coordinates. I have tried several different ways. I do not arrive at these coordinates.
I am trying to graph the following:
y>3/2x-5/2
(sign > is greater then or equal to)
The way I understand it I would start on the Y axis at 0 and go down 5 and then to the right 2 for the -5/2. Then I would go up 3 and to the right 2 for the 3/2. The book shows(0,3/2) and (2,-7/2) as the coordinates. I have tried several different ways. I do not arrive at these coordinates.
Answers
Answered by
Reiny
No, the first step in your process is wrong.
The -5/2 IS the y-intercept and -5/2 is -2.5
So you are at (0,-2.5), now go up 3 and to the right 2, for the second point (2,0.5)
In fraction form these points should be
(0, -5/2) and (2,1/2)
Join them with a solid line, and shade in the region above the line.
Unless you typed the inequation incorrectly, the book answers you supplied are wrong.
The -5/2 IS the y-intercept and -5/2 is -2.5
So you are at (0,-2.5), now go up 3 and to the right 2, for the second point (2,0.5)
In fraction form these points should be
(0, -5/2) and (2,1/2)
Join them with a solid line, and shade in the region above the line.
Unless you typed the inequation incorrectly, the book answers you supplied are wrong.
Answered by
G
Thank you for your response.
This is a rough graph with the exact coordinates stated in the book.
Y
-
-
.(0,3/2)
-
0 ' ' ' ' 'X
-
-
- .(2,-7/2)
-
-
This is a rough graph with the exact coordinates stated in the book.
Y
-
-
.(0,3/2)
-
0 ' ' ' ' 'X
-
-
- .(2,-7/2)
-
-
Answered by
Reiny
My point was and my point still is that the book is wrong if these points are supposed to satisfy the "equation" part of the inequality.
The points I gave you DO satisfy the equation part of the inequality, thus they establish the boundary of the region you want.
The points I gave you DO satisfy the equation part of the inequality, thus they establish the boundary of the region you want.
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