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Look at the system of inequalities. $\newline$ $y \leq -2x + 8$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ The solution set is the triangular region where all the inequalities are true. $\newline$ What are the vertices of that triangular region? $\newline$ (,) $\newline$ (,) $\newline$ (,)

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Find the focus or directrix of a parabola

Full solution

Q. Look at the system of inequalities.

\newline

y \leq -2x + 8

\newline

x \geq 0

\newline

y \geq 0

\newline

The solution set is the triangular region where all the inequalities are true.

\newline

What are the vertices of that triangular region?

\newline

(____,____)

\newline

(____,____)

\newline

(____,____)

Find Intersection of Inequalities: First, let's find the intersection of $y \leq -2x + 8$ and $x \geq 0$ . $\newline$ Set $x = 0$ in the first inequality: $y \leq 8$ . $\newline$ So one vertex is at $(0, 8)$ .
Identify Vertices: Next, find the intersection of $y \leq -2x + 8$ and $y \geq 0$ . Set $y = 0$ in the first inequality: $0 \leq -2x + 8$ . Solve for $x$ : $2x = 8$ ; $x = 4$ . So another vertex is at $(4, 0)$ .
Check Vertex Validity: Now, find the intersection of $x \geq 0$ and $y \geq 0$ . The intersection is the origin, so the third vertex is at $(0, 0)$ .
Check Vertex Validity: Now, find the intersection of $x \geq 0$ and $y \geq 0$ . The intersection is the origin, so the third vertex is at $(0, 0)$ . Finally, we need to check if all three vertices satisfy all the inequalities. $(0, 8)$ and $(4, 0)$ are on the boundary lines of the inequalities, and $(0, 0)$ is within the first quadrant where $x$ and $y$ are non-negative.

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Question

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1

\newline

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