document.write( "Question 585129: I am trying to graph the function:y=-4(x+3)+2. The graph looks like the square function graph(intersecting at (0,0) and opening up with points that appear to be (2,4) and (-2,4)but it is identified as y=f(x), so, I,m just confused do I solve it like I would if it is a square function graph, If not where do I start? \n" ); document.write( "
Algebra.Com's Answer #373145 by lwsshak3(11628)\"\" \"About 
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I am trying to graph the function:y=-4(x+3)+2. The graph looks like the square function graph(intersecting at (0,0) and opening up with points that appear to be (2,4) and (-2,4)but it is identified as y=f(x), so, I,m just confused do I solve it like I would if it is a square function graph, If not where do I start?
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\n" ); document.write( "The first thing you do to solve and graph given equation is to clear the parenthesis and rewrite the equation in standard form.
\n" ); document.write( "y=-4(x+3)+2
\n" ); document.write( "Clear parenthesis
\n" ); document.write( "y=-4x-12+2
\n" ); document.write( "rewrite in standard form
\n" ); document.write( "y=4x-10
\n" ); document.write( "This is an equation of a straight line with slope m=4, and y-intercept=-10.
\n" ); document.write( "To graph this line, you can plot points, but a better way is to start with the y-intercept at (0,-10), then from this point move 4 unit up and one unit to the right (rise/run) to establish the second point of the straight line. These two points are on the straight line.
\n" ); document.write( "Note: First degree equations like this one graph as straight lines and second degree equations
\n" ); document.write( "graph as parabolas. \n" ); document.write( "
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