Question

1. (0, 1)
2. graph each inequality in the coordinate plane.
3. ( y geq 2x )
4. ( y < x - 2 )
5. what inequality is shown by the graph?

graph of a line with y-intercept at (0, 1) and x-intercept at (2, 0), dashed line, shaded region below the line
inequalities in two variables

Explanation:

To determine the inequality shown by the graph, we analyze the line and the shaded region (or the side of the line the solution is on).

Step 1: Find the equation of the line

The line passes through two points: let's identify them from the graph. From the graph, we can see the line passes through \((0, 1)\) (the y - intercept) and \((2, 0)\) (the x - intercept).

The slope \(m\) of a line passing through \((x_1,y_1)\) and \((x_2,y_2)\) is given by \(m=\frac{y_2 - y_1}{x_2 - x_1}\).

Substituting \((x_1,y_1)=(0,1)\) and \((x_2,y_2)=(2,0)\):
\(m=\frac{0 - 1}{2 - 0}=\frac{-1}{2}=-\frac{1}{2}\)

Using the slope - intercept form of a line \(y = mx + b\), where \(b\) is the y - intercept. Here, \(b = 1\) (since the line crosses the y - axis at \((0,1)\)) and \(m=-\frac{1}{2}\). So the equation of the line is \(y=-\frac{1}{2}x + 1\).

Step 2: Determine the inequality symbol

• The line is a dashed line (from the graph, we can see it's a dashed line, which means the inequality is strict, either \(<\) or \(>\), not \(\leq\) or \(\geq\)).
• Now, we need to determine which side of the line is shaded. Let's test a point not on the line. A common test point is the origin \((0,0)\).

Substitute \(x = 0\) and \(y = 0\) into the inequality. Let's assume the inequality is \(y<-\frac{1}{2}x + 1\) (we can also test \(y>-\frac{1}{2}x + 1\)).

For \(y<-\frac{1}{2}x + 1\):
Left - hand side (LHS): \(y = 0\)
Right - hand side (RHS): \(-\frac{1}{2}(0)+1=1\)

Since \(0<1\) is true, the origin \((0,0)\) satisfies the inequality \(y<-\frac{1}{2}x + 1\). Looking at the graph, if the origin is in the solution region (the side of the line that is part of the inequality), and the line is dashed, the inequality is \(y<-\frac{1}{2}x + 1\) (we can also rewrite it in standard form: \(\frac{1}{2}x+y<1\) or \(x + 2y<2\), but the slope - intercept form is more intuitive here).

Answer:

The inequality shown by the graph is \(y<-\frac{1}{2}x + 1\) (or equivalent forms like \(x + 2y<2\)).