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Look at the system of inequalities. $\newline$ $y \leq -\frac{1}{3}x + 2$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 -\frac{1}{3}x + 2

\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 with $x=0$ : First, let's find the intersection of $y \leq -\frac{1}{3}x + 2$ and $x \geq 0$ . Set $x = 0$ in the first inequality. $y \leq -\frac{1}{3}(0) + 2$ $y \leq 2$ So, one vertex is at $(0, 2)$ .
Find Intersection with y= $0$ : Next, find the intersection of $y \leq -\frac{1}{3}x + 2$ and $y \geq 0$ . Set $y = 0$ in the first inequality. $0 \leq -\frac{1}{3}x + 2$ Multiply both sides by $-3$ to get rid of the fraction. $0 \geq x - 6$ Add $6$ to both sides. $6 \geq x$ So, another vertex is at $(6, 0)$ .
Find Origin Intersection: Finally, find the intersection of $x \geq 0$ and $y \geq 0$ . This is simply the origin, where both $x$ and $y$ are zero. So, the last vertex is at $(0, 0)$ .

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