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Look at the system of inequalities. $\newline$ $y \leq -\frac{1}{2}x + 2$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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}{2}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-Axis: First, let's find the intersection of $y = -\frac{1}{2}x + 2$ and the x-axis ( $y = 0$ ). $\newline$ Set $y = 0$ in the equation $y = -\frac{1}{2}x + 2$ and solve for $x$ . $\newline$ $0 = -\frac{1}{2}x + 2$ $\newline$ $\frac{1}{2}x = 2$ $\newline$ $x = 4$ $\newline$ So, the intersection point on the x-axis is $(4, 0)$ .
Find Intersection with Y-Axis: Next, find the intersection of $y = -\frac{1}{2}x + 2$ and the y-axis ( $x = 0$ ). $\newline$ Set $x = 0$ in the equation $y = -\frac{1}{2}x + 2$ and solve for $y$ . $\newline$ $y = -\frac{1}{2}(0) + 2$ $\newline$ $y = 2$ $\newline$ So, the intersection point on the y-axis is $(0, 2)$ .
Identify Origin as Vertex: Lastly, since $x \geq 0$ and $y \geq 0$ , the origin $(0, 0)$ is also a vertex of the triangular region.

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