Question

graph the inequality on the axes below.
$-x + 2y > 4$

Explanation:

Step1: Rewrite the inequality in slope - intercept form

We start with the inequality \(-x + 2y>4\). To get it in the form \(y = mx + b\) (slope - intercept form), we solve for \(y\).
First, add \(x\) to both sides of the inequality: \(2y>x + 4\).
Then, divide each term by 2: \(y>\frac{1}{2}x+2\).

Step2: Graph the boundary line

The boundary line is \(y=\frac{1}{2}x + 2\). Since the inequality is \(y>\frac{1}{2}x + 2\) (not \(y\geq\frac{1}{2}x + 2\)), the boundary line should be a dashed line.

• To find the \(y\) - intercept: When \(x = 0\), \(y=\frac{1}{2}(0)+2=2\). So the line passes through the point \((0,2)\).
• To find another point, we can use the slope. The slope \(m=\frac{1}{2}\), which means for every 2 units we move to the right (increase in \(x\) by 2), we move up 1 unit (increase in \(y\) by 1). So from \((0,2)\), if we move \(x = 2\), then \(y=\frac{1}{2}(2)+2=1 + 2=3\). So the line also passes through the point \((2,3)\). We can plot these two points \((0,2)\) and \((2,3)\) and draw a dashed line through them.

Step3: Shade the region

We test a point that is not on the line to see which side of the line to shade. A common test point is the origin \((0,0)\).
Substitute \(x = 0\) and \(y = 0\) into the inequality \(y>\frac{1}{2}x+2\):
Left - hand side: \(y = 0\)
Right - hand side: \(\frac{1}{2}(0)+2=2\)
Since \(0>2\) is false, we shade the region that does not contain the origin. In other words, we shade the region above the dashed line \(y=\frac{1}{2}x + 2\).

(Note: Since the problem asks to graph the inequality, the final answer is the graph with a dashed line \(y = \frac{1}{2}x+2\) and the region above the line shaded. If we were to describe the key steps of the graphing process, the above steps are the main ones.)

Answer:

The graph has a dashed line \(y=\frac{1}{2}x + 2\) (passing through \((0,2)\) and \((2,3)\)) and the region above the line is shaded.