document.write( "Question 965953: graph
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Algebra.Com's Answer #590389 by jim_thompson5910(35256)\"\" \"About 
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Solved by pluggable solver: Graphing Linear Equations
In order to graph \"y=9%2Ax-3\" 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=0
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\n" ); document.write( " \"y=9%2A%280%29-3\"
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\n" ); document.write( " \"y=0-3\" Multiply
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\n" ); document.write( " \"y=-3\" Add
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\n" ); document.write( " So here's one point (0,-3)
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\n" ); document.write( " Now lets find another point
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\n" ); document.write( " Plug in x=1
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\n" ); document.write( " \"y=9%2A%281%29-3\"
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\n" ); document.write( " \"y=9-3\" Multiply
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\n" ); document.write( " \"y=6\" Add
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\n" ); document.write( " So here's another point (1,6). 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=9%2Ax-3\" through the points (0,-3) and (1,6)
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\n" ); document.write( " So from the graph we can see that the slope is \"9%2F1\" (which tells us that in order to go from point to point we have to start at one point and go up 9 units and to the right 1 units to get to the next point) the y-intercept is (0,\"-3\")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=-3\" this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,\"-3\").
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\n" ); document.write( " So we have one point (0,\"-3\")
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\n" ); document.write( " Now since the slope is \"9%2F1\", this means that in order to go from point to point we can use the slope to do so. So starting at (0,\"-3\"), we can go up 9 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=9%2Ax-3\"
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\n" ); document.write( " So this is the graph of \"y=9%2Ax-3\" through the points (0,-3) and (1,6)
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