Question 169716


In order to graph this equation, we only need two points to create a straight line





--------------------------------Let's find the first point--------------------------------


{{{y=-2x-1}}} Start with the given equation





{{{y=-2(0)-1}}} Plug in {{{x=0}}} (you can start with any number, but zero is the easiest number to work with since anything multiplied by zero is zero)





{{{y=0-1}}} Multiply {{{-2}}} and {{{0}}} to get {{{0}}}





{{{y=-1}}} Subtract 




So when {{{x=0}}}, we have the value {{{y=-1}}}. This means we have the first point *[Tex \LARGE \left(0,-1\right)]





--------------------------------Let's find the second point--------------------------------


{{{y=-2x-1}}} Start with the given equation





{{{y=-2(1)-1}}} Plug in {{{x=1}}} 





{{{y=-2-1}}} Multiply {{{-2}}} and {{{1}}} to get {{{-2}}}





{{{y=-3}}} Subtract 




So when {{{x=1}}}, we have the value {{{y=-3}}}. This means we have the second point *[Tex \LARGE \left(1,-3\right)]





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So we have the two points: *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(1,-3\right)]



Now plot these two points on a coordinate system


{{{drawing(500,500,-9.5,10.5,-9.5,10.5,
graph(500,500,-9.5,10.5,-9.5,10.5,0),
grid(1),
circle(0,-1,0.1),
circle(0,-1,0.12),
circle(0,-1,0.15),
circle(1,-3,0.1),
circle(1,-3,0.12),
circle(1,-3,0.15)
) }}}




Now draw a straight line through the two points. This line is the graph of {{{y=-2x-1}}}


{{{drawing(500,500,-9.5,10.5,-9.5,10.5,
graph(500,500,-9.5,10.5,-9.5,10.5,-2x-1),
grid(1),
circle(0,-1,0.1),
circle(0,-1,0.12),
circle(0,-1,0.15),
circle(1,-3,0.1),
circle(1,-3,0.12),
circle(1,-3,0.15)
) }}} Graph of {{{y=-2x-1}}} through the two points *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(1,-3\right)]