Question

using the graphing calculator to include a solution
mr. hernandez plotted the point (1, 1) on han’s 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?
\\(\circ\\) $y > 2x + 1$
\\(\circ\\) $y < 2x - 1$
\\(\circ\\) $y \geq 2x + 1$
\\(\circ\\) $y \leq 2x - 1$
(graph: a coordinate plane with the line $y \leq \frac{1}{2}x + 2$ graphed, shaded region, and point (1,1) plotted)

Explanation:

Step1: Substitute (1,1) into each inequality

For option 1: \( y > 2x + 1 \)
Substitute \( x = 1 \), \( y = 1 \): \( 1 > 2(1) + 1 \) → \( 1 > 3 \), which is false.

Step2: Check option 2: \( y < 2x - 1 \)

Substitute \( x = 1 \), \( y = 1 \): \( 1 < 2(1) - 1 \) → \( 1 < 1 \), which is false.

Step3: Check option 3: \( y \geq 2x + 1 \)

Substitute \( x = 1 \), \( y = 1 \): \( 1 \geq 2(1) + 1 \) → \( 1 \geq 3 \), which is false? Wait, no, wait, 2(1)+1 is 3? Wait, no, 2x+1 when x=1 is 3? Wait, no, 2*1 +1=3? Wait, 1 is not greater or equal to 3. Wait, maybe I miscalculated. Wait, the options: let's recheck.

Wait, the options are:

1. \( y > 2x + 1 \)

1. \( y < 2x - 1 \)

1. \( y \geq 2x + 1 \)

1. \( y \leq 2x - 1 \)

Wait, when x=1, 2x+1=3, 2x-1=1.

So for option 1: 1 > 3? No.

Option 2: 1 < 1? No.

Option 3: 1 ≥ 3? No. Wait, that can't be. Wait, maybe I made a mistake. Wait, 2x -1 when x=1 is 1. So option 4: \( y \leq 2x - 1 \): 1 ≤ 1? Yes! Oh! Wait, I misread option 4. Let's recheck:

Option 4: \( y \leq 2x - 1 \)

Substitute x=1, y=1: 1 ≤ 2(1) -1 → 1 ≤ 1, which is true.

Wait, earlier I thought option 3 was \( y \geq 2x +1 \), but let's re-express:

Wait, 2x -1 when x=1 is 1. So option 4: \( y \leq 2x -1 \) → 1 ≤ 1, which is true.

Wait, let's recheck each option:

1. \( y > 2x + 1 \): x=1, y=1: 1 > 3? No.

1. \( y < 2x - 1 \): x=1, y=1: 1 < 1? No.

1. \( y \geq 2x + 1 \): x=1, y=1: 1 ≥ 3? No.

1. \( y \leq 2x - 1 \): x=1, y=1: 1 ≤ 1? Yes.

Ah, I see, I misread option 4. So the correct one is option 4: \( y \leq 2x - 1 \). Wait, but let's confirm.

Wait, the point (1,1). Let's check each inequality:

1. \( y > 2x + 1 \): 1 > 2(1) +1 → 1 > 3? No.

1. \( y < 2x - 1 \): 1 < 2(1) -1 → 1 < 1? No.

1. \( y \geq 2x + 1 \): 1 ≥ 3? No.

1. \( y \leq 2x - 1 \): 1 ≤ 1? Yes. So the correct option is the fourth one: \( y \leq 2x - 1 \). Wait, but the options are labeled as:

The options are:

○ \( y > 2x + 1 \)

○ \( y < 2x - 1 \)

○ \( y \geq 2x + 1 \)

○ \( y \leq 2x - 1 \)

So the fourth option is \( y \leq 2x - 1 \), which when x=1, y=1, satisfies 1 ≤ 1. So that's the correct one.

Answer:

The correct option is the fourth one: \( y \leq 2x - 1 \) (the option labeled as \( y \leq 2x - 1 \)).