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Look at the system of inequalities. $\newline$ $y \leq -\frac{1}{2}x + 4$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 -\frac{1}{2}x + 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 with $x=0$ : First, let's find the intersection of $y \leq -\frac{1}{2}x + 4$ and $x \geq 0$ . $\newline$ Set $x = 0$ in the first inequality to find the y-intercept. $\newline$ $y \leq -\frac{1}{2}(0) + 4$ $\newline$ $y \leq 4$ $\newline$ So, one vertex is at $(0, 4)$ .
Find Intersection with y= $0$ : Now, let's find the intersection of $y \leq -\frac{1}{2}x + 4$ and $y \geq 0$ . Set $y = 0$ in the first inequality to find the x-intercept. $0 \leq -\frac{1}{2}x + 4$ $-4 \leq -\frac{1}{2}x$ $8 \geq x$ So, another vertex is at $(8, 0)$ .
Find Origin Intersection: Lastly, we need to 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 third vertex is at $(0, 0)$ .

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