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Look at the system of inequalities. $\newline$ $y \leq -2x + 4$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 + 4

\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 $y \leq -2x + 4$ and $x \geq 0$ . $\newline$ Set $x = 0$ in the first inequality. $\newline$ $y \leq -2(0) + 4$ $\newline$ $y \leq 4$ $\newline$ So one vertex is at $(0, 4)$ .
Vertex at $(0, 4)$ : Now, let's find the intersection of $y \leq -2x + 4$ and $y \geq 0$ . $\newline$ Set $y = 0$ in the first inequality. $\newline$ $0 \leq -2x + 4$ $\newline$ $2x \leq 4$ $\newline$ $x \leq 2$ $\newline$ But since $x \geq 0$ , the intersection is at $x = 2$ . $\newline$ Substitute $x = 2$ into $y \leq -2x + 4$ $0$ to find the y-coordinate. $\newline$ $y \leq -2x + 4$ $1$ $\newline$ $y \leq -2x + 4$ $2$ $\newline$ $y = 0$ $\newline$ So another vertex is at $y \leq -2x + 4$ $4$ .
Vertex at $(2, 0)$ : Lastly, we need to find the intersection of $x \geq 0$ and $y \geq 0$ , which is simply the origin. $\newline$ So the third vertex is at $(0, 0)$ .

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