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How do you graph #y<=-x+2# on the coordinate plane?

Answer 1

Julia Adams

See a solution process below:

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

For: #x = 0#

#y = -0 + 2#

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

For: #x = 2#

#y = -2 + 2#

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

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. The boundary line will be solid because the inequality operator contains an "or equal to" clause.

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

Now, we can shade the left side of the line.

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

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Answer 2

Eli Assouline

To graph the inequality ( y \leq -x + 2 ) on the coordinate plane, first graph the line ( y = -x + 2 ) as a solid line. Then, shade the region below the line, including the line itself, since it is a less than or equal to inequality.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you graph #y&lt;=-x+2# on the coordinate plane?

How do you graph #y<=-x+2# on the coordinate plane?