Question

using the graphing calculator to include a solution
mr. hernandez plotted the point (1, 1) on hans graph
of ( y leq \frac{1}{2}x + 2 ). he instructed han to add a second
inequality to the graph that would include the solution
(1, 1).
which equation could miguel write?
( \bigcirc y>2x + 1 )
( \bigcirc y<2x - 1 )
( \bigcirc ygeq2x + 1 )
( \bigcirc yleq2x - 1 )

Explanation:

Step1: Substitute \(x = 1\) and \(y = 1\) into each inequality

• For \(y>2x + 1\):

Substitute \(x = 1\) and \(y = 1\) into \(y>2x + 1\). We get \(1>2\times1+1\), i.e., \(1>3\) (False).

• For \(y<2x - 1\):

Substitute \(x = 1\) and \(y = 1\) into \(y<2x - 1\). We get \(1<2\times1-1\), i.e., \(1<1\) (False).

• For \(y\geq2x + 1\):

Substitute \(x = 1\) and \(y = 1\) into \(y\geq2x + 1\). We get \(1\geq2\times1+1\), i.e., \(1\geq3\) (False).

• For \(y\leq2x - 1\):

Substitute \(x = 1\) and \(y = 1\) into \(y\leq2x - 1\). We get \(1\leq2\times1-1\), i.e., \(1\leq1\) (True).

Answer:

\(y\leq2x - 1\)