How do you graph the equation #y=-3/7x+2#?

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Answer 1 · 2017-10-17T21:52:57

See a solution process below:

This equation is in slope intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value.

#y = color(red)(-3/7)x + color(blue)(2)#

Therefore:

The #y#-intercept is: #color(blue)(2)# or #(0, color(blue)(2))#

The slope is: #color(red)(m = -3/7)#

Slope is rise over run. So the line will go down #3# units while it goes to the right #7# units.

We can plot the #y#-intercept as:

graph{(x^2+(y-2)^2-0.025)=0}

We can plot the next point by going down #3# units and to the right #7# units which is at: #(7, -1)#

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We can now draw a line through the two points to graph the line:

graph{(y + (3/7)x - 2)((x-7)^2+(y+1)^2-0.025)(x^2+(y-2)^2-0.025)=0}