Question

on a piece of paper, graph $y + 2 \leq \frac{1}{4}x - 1$. then determine which answer choice matches the graph you drew.
a
b
c
d
\bigcirc a. graph a
\bigcirc b. graph b
\bigcirc c. graph c
\bigcirc d. graph d

Explanation:

Step1: Rewrite the inequality

First, we rewrite the given inequality \( y + 2\leq\frac{1}{4}x - 1 \) in slope - intercept form (\(y=mx + b\)) by solving for \(y\).
Subtract 2 from both sides of the inequality:
\(y\leq\frac{1}{4}x-1 - 2\)
\(y\leq\frac{1}{4}x-3\)

Step2: Analyze the boundary line

The boundary line of the inequality \(y\leq\frac{1}{4}x - 3\) is the line \(y=\frac{1}{4}x-3\).

• Slope and y - intercept: The slope \(m = \frac{1}{4}\) and the y - intercept \(b=-3\) (so the line passes through the point \((0,-3)\)).
• Type of line: Since the inequality is \(\leq\), the boundary line should be a solid line (because the points on the line are included in the solution set).

Step3: Determine the region to shade

To determine which side of the line to shade, we can use a test point. A common test point is \((0,0)\) (as long as it is not on the boundary line).
Substitute \(x = 0\) and \(y = 0\) into the inequality \(y\leq\frac{1}{4}x-3\):
\(0\leq\frac{1}{4}(0)-3\)
\(0\leq - 3\), which is false. So we shade the region that does not contain the point \((0,0)\).

Now let's analyze the graphs:

• Graph A: The boundary line is solid, and the shading is below the line (since when we test \((0,0)\) it is not in the shaded region). The line passes through \((0, - 3)\) and has a slope of \(\frac{1}{4}\) (we can check the slope between \((0,-3)\) and \((4,-2)\): \(m=\frac{-2 - (-3)}{4-0}=\frac{1}{4}\), which matches the slope of our line \(y = \frac{1}{4}x-3\)).
• Graph B: The shading is above the line, which would be the case if the inequality was \(y\geq\frac{1}{4}x - 3\), so this is incorrect.
• Graph C and D: The boundary lines are dashed, but our boundary line should be solid (because of the \(\leq\) inequality), so these are incorrect.

Answer:

A. Graph A