Graph y=2/3x-5 using the slope and the vertical intercept

jamsey777

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Nov 29, 2007
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I am so confused! I am working on a problem that goes as follows....

Graph y=2/3x-5 using the slope and the vertical intercept.

Graphing is clearly not my strongest point.....any help would be so appreciated!! :D
 
jamsey777 said:
Graph y=2/3x-5 using the slope and the vertical intercept.
In "y = mx + b", the slope is "m" and the vertical intercept is at (0, b). So where is the y-intercept for y = (2/3)x - 5? What is the slope?

What is the meaning of slope? Given one point on the graph, how does slope allow you to "count" up and over to the next point? :wink:

Eliz.
 
jamsey777 said:
IGraphing is clearly not my strongest point.
You better get used to graphing because you will see it in every math course beginning with prealgebra and nonestop from there on, HONESTLY. What about when you face graphing trig functions? How about graphing exponential functions, rational functions and the list goes on and on???
 
In the form

y = mx + b

the constant term, b, is where the graph crosses the y-axis.

You have

y = (2/3)x - 5

This graph crosses the y-axis at -5. Start by making a dot there. Then use the slope. A slope of 2/3 means that when you go UP 2 units, you move to the RIGHT 3 units (rise over run). Start at the dot you made on the y-axis. Go up 2 units and to the right 3 units. Make another dot there--it should be at (3, -3). Then draw the line through the two dots.

This line is the graph of the equation y = (2/3)x - 5.
 
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