IXL - Linear programming (Algebra 2 practice) - Free Printable

Let’s find the vertices of the triangular region formed by these inequalities:

We are given:
1. \( y \leq -\frac{1}{2}x + 4 \)
2. \( x \geq 0 \)
3. \( y \geq 0 \)

The solution is a triangle bounded by these three lines. The vertices (corners) of the triangle occur where the boundary lines intersect each other.

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Step 1: Find where \( x = 0 \) and \( y = 0 \) meet.

That’s the origin:
→ Point A: \( (0, 0) \)

This point satisfies all three inequalities:
- \( y = 0 \leq -\frac{1}{2}(0) + 4 = 4 \) → true
- \( x = 0 \geq 0 \) → true
- \( y = 0 \geq 0 \) → true

So (0, 0) is one vertex.

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Step 2: Find where \( x = 0 \) meets the line \( y = -\frac{1}{2}x + 4 \)

Plug in \( x = 0 \):

\( y = -\frac{1}{2}(0) + 4 = 4 \)

→ Point B: \( (0, 4) \)

Check if it satisfies all inequalities:
- \( y = 4 \leq 4 \) → true
- \( x = 0 \geq 0 \) → true
- \( y = 4 \geq 0 \) → true

Good — this is another vertex.

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Step 3: Find where \( y = 0 \) meets the line \( y = -\frac{1}{2}x + 4 \)

Set \( y = 0 \):

\( 0 = -\frac{1}{2}x + 4 \)

Solve for x:

Add \( \frac{1}{2}x \) to both sides:

\( \frac{1}{2}x = 4 \)

Multiply both sides by 2:

\( x = 8 \)

→ Point C: \( (8, 0) \)

Check inequalities:
- \( y = 0 \leq -\frac{1}{2}(8) + 4 = -4 + 4 = 0 \) → true
- \( x = 8 \geq 0 \) → true
- \( y = 0 \geq 0 \) → true

Perfect — third vertex.

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So the three vertices are:
- (0, 0)
- (0, 4)
- (8, 0)

These match the black dots shown on the graph too!

Final Answer:
(0, 0)
(0, 4)
(8, 0)