How do you graph #y=3/2x-4# using the slope and intercept?

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Answer 1 · 2017-10-16T18:05:18

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)(3/2)x - color(blue)(4)#

Therefore:

The #y#-intercept is: #color(blue)(-4)# or #(0, color(blue)(-4))#

The slope is: #color(red)(m = 3/2)#

Slope is rise over rub. So the line will go up #3# units while it goes to the right #2# units.

We can plot the #y#-intercept as:

graph{(x^2+(y+4)^2-0.025)=0}

We can plot the next point by going up #3# units and to the right #2# units which is at: #(2, -1)#

enter image source here

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

graph{(x^2+(y+4)^2-0.025)((x-2)^2+(y+1)^2-0.025)(y-(3/2)x+4)=0}