document.write( "Question 315642: graph an equation of the form y = mx + b\r
\n" ); document.write( "\n" ); document.write( "y = -2x + 2 \n" ); document.write( "

Algebra.Com's Answer #226377 by moshiz08(60) $\"\"$ $\"About$

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$\"y+=+-2x+%2B+2\"$ \r
\n" ); document.write( "\n" ); document.write( "Plug in some values for x to get an idea of what points are on the curve. For example, when x=0, we get y=2. When x=1, we get y=0. We can plot these points on a graph. The function that you are graphing is linear, so as long as we know two points, we can make a straight line going through them for the graph. It looks like this: \r
\n" ); document.write( "\n" ); document.write( " $\"+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+-2x%2B2%29+\"$ \r
\n" ); document.write( "\n" ); document.write( "Some key points to notice: this is in slope-intercept form. \r
\n" ); document.write( "\n" ); document.write( "The slope is -2. Since the slope is negative, when you go to the right, the graph will go down. In particular, when you go right 1 unit, you go down 2 units to stay on the curve. \r
\n" ); document.write( "\n" ); document.write( "The +2 part gives you the y-intercept. That is, the graph crosses the y-axis at y=2. \n" ); document.write( "

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