How do you graph y<3x-3/4 on the coordinate plane?

Mar 19, 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: $x = 0$

$y = \left(3 \cdot 0\right) - \frac{3}{4}$

2y = 0 - 3/4#

$y = - \frac{3}{4}$ or $\left(0 , - \frac{3}{4}\right)$

For: $x = 1$

$y = \left(3 \cdot 1\right) - \frac{3}{4}$

$y = 3 - \frac{3}{4}$

$y = \left(\frac{4}{4} \cdot 3\right) - \frac{3}{4}$

$y = \frac{12}{4} - \frac{3}{4}$

$y = \frac{9}{4}$ or $\left(1 , \frac{9}{4}\right)$

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+(3/4))^2-0.035)((x-1)^2+(y-(9/4))^2-0.035)(y-3x+(3/4))=0 [-10, 10, -5, 5]}

Now, we can shade the right side of the line to represent the inequality.

The boundary line will be changed to a dashed line because the inequality operator does not contain an "or equal to" clause.

graph{(y-3x+(3/4)) < 0 [-10, 10, -5, 5]}