SOLUTION: 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

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?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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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The first thing you do to solve and graph given equation is to clear the parenthesis and rewrite the equation in standard form.
y=-4(x+3)+2
Clear parenthesis
y=-4x-12+2
rewrite in standard form
y=4x-10
This is an equation of a straight line with slope m=4, and y-intercept=-10.
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.
Note: First degree equations like this one graph as straight lines and second degree equations
graph as parabolas.
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