Given the inequality:
[tex]y<\frac{1}{3}x+1[/tex]
Let's select the graph which correctly shows the boundary line of the inequality.
Using the inequality given, let's find random points on which the boundary line passes through.
Rewrite the inequality as an equation:
[tex]y=\frac{1}{3}x+1[/tex]
Find y when x = 0:
[tex]\begin{gathered} y=\frac{1}{3}(0)+1 \\ y=0+1 \\ y=1 \end{gathered}[/tex]
Find y when x = 3:
[tex]\begin{gathered} y=\frac{1}{3}(3)+1 \\ y=1+1 \\ y=2 \end{gathered}[/tex]
Find y when x = -3:
[tex]\begin{gathered} y=\frac{1}{3}(-3)+1 \\ y=-1+1 \\ y=0 \end{gathered}[/tex]
Therefore, we have the points:
(x, y) ==> (0, 1), (3, 2), (-3, 0)
Since y is less than the expression, the line will be a dashed line.
Now, let's check the graph which has a dashed line that passes these points.
Therefore, the correct graph is:
ANSWER:
Graph C.
