How do you graph #y=3x-2# using slope intercept form?

1 Answer
Feb 17, 2018

Answer:

See a solution process below:

Explanation:

First, 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)x - color(blue)(2)#

Or

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

Therefore, we know the slope is: #color(red)(m = 3)#

And the #y#-intercept is: #color(blue)(b = -2)# or #(0, color(blue)(-2))#

We can start graphing this equation by plotting the #y#-intercept:

graph{(x^2 + (y+2)^2 - 0.025) = 0 [-10, 10, -5, 5]}

Slope is defined as #"rise"/"run"#, or the amount the #y# value changes compared to the #x# value.

The slope for this equation is #m = 3# or #m = 1#.

Therefore for each change in #y# of #3#, #x# changes by #1#.

We can now plot another point using this information:

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Now, we can draw a straight line through the two points to graph the equation:

graph{(y - 3x +2)(x^2 + (y+2)^2 - 0.025)((x - 1)^2 + (y - 1)^2 - 0.025) = 0 [-10, 10, -5, 5]}