document.write( "Question 118020: Graph the following equation.\r
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Algebra.Com's Answer #86063 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
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Solved by pluggable solver: Graphing Linear Equations
In order to graph \"y=3%2Ax%2B1\" we only need to plug in two points to draw the line
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\n" ); document.write( " So lets plug in some points
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\n" ); document.write( " Plug in x=-3
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\n" ); document.write( " \"y=3%2A%28-3%29%2B1\"
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\n" ); document.write( " \"y=-9%2B1\" Multiply
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\n" ); document.write( " \"y=-8\" Add
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\n" ); document.write( " So here's one point (-3,-8)
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\n" ); document.write( " Now lets find another point
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\n" ); document.write( " Plug in x=-2
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\n" ); document.write( " \"y=3%2A%28-2%29%2B1\"
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\n" ); document.write( " \"y=-6%2B1\" Multiply
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\n" ); document.write( " \"y=-5\" Add
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\n" ); document.write( " So here's another point (-2,-5). Add this to our graph
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\n" ); document.write( " Now draw a line through these points
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\n" ); document.write( " So this is the graph of \"y=3%2Ax%2B1\" through the points (-3,-8) and (-2,-5)
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\n" ); document.write( " So from the graph we can see that the slope is \"3%2F1\" (which tells us that in order to go from point to point we have to start at one point and go up 3 units and to the right 1 units to get to the next point) the y-intercept is (0,\"1\")and the x-intercept is (\"-0.333333333333333\",0) ,or (\"-1%2F3\",0)
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\n" ); document.write( " We could graph this equation another way. Since \"b=1\" this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,\"1\").
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\n" ); document.write( " So we have one point (0,\"1\")
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\n" ); document.write( " Now since the slope is \"3%2F1\", this means that in order to go from point to point we can use the slope to do so. So starting at (0,\"1\"), we can go up 3 units
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\n" ); document.write( " and to the right 1 units to get to our next point
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\n" ); document.write( " Now draw a line through those points to graph \"y=3%2Ax%2B1\"
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\n" ); document.write( " So this is the graph of \"y=3%2Ax%2B1\" through the points (0,1) and (1,4)
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