How do you graph the inequality #y<4(x)+1#?

1 Answer
Jan 12, 2018

Answer:

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 = (4 * 0) + 1#

#y = 0 + 1#

#y = 1# or #(0, 1)#

For: #x = 2#

#y = (4 * 2) + 1#

#y = 8 + 1#

#y = 9# or #(2, 9)#

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-1)^2-0.125)((x-2)^2+(y-9)^2-0.125)(y-4x - 1)=0 [-20, 20, -10, 10]}

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

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

graph{(y- 4x - 1) < 0 [-20, 20, -10, 10]}