How do you graph #y-2<3x#?

1 Answer
smendyka
Jan 3, 2018

See a solution process below:

Explanation:

First, solve for two points as an equation instead of an inequality to find the boundary line for the inequality.

For: #x = 0#

#y - 2 = 3 xx 0#

#y - 2 = 0#

#y - 2 + color(red)(2) = 0 + color(red)(2)#

#y - 0 = 2#

#y = 2# or #(0, 2)#

For: #x = 1#

#y - 2 = 3 xx 1#

#y - 2 = 3#

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

#y - 0 = 5#

#y = 5# or #(1, 5)#

We can now graph the two points on the coordinate plane and draw a line through the points to mark the boundary of the inequality.

graph{(x^2+(y-2)^2-0.05)((x-1)^2+(y-5)^2-0.05)(y-3x-2)=0 [-16, 16, -8, 8]}

Now, we can shade the right side of the line. The boundary line will be dashed because the inequality operator does not contain an "or equal to" clause.

graph{(y-3x-2)<0 [-16, 16, -8, 8]}

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