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

\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 $y = -\frac{1}{2}x + 3$ and the x-axis ( $y=0$ ). $\newline$ Set $y$ to $0$ and solve for $x$ : $0 = -\frac{1}{2}x + 3$ .
Solve for $X$ : Multiply both sides by $2$ to get rid of the fraction: $0 = -x + 6$ .
Intersection at $(6,0)$ : Add $x$ to both sides to solve for $x$ : $x = 6$ . So, the intersection with the $x$ -axis is at $(6,0)$ .
Find Intersection with Y-Axis: Now, let's find the intersection of $y = -\frac{1}{2}x + 3$ and the y-axis ( $x=0$ ). $\newline$ Set $x$ to $0$ and solve for $y$ : $y = -\frac{1}{2}(0) + 3$ .
Solve for Y: Simplify to find $y$ : $y = 0 + 3$ . $\newline$ So, the intersection with the y-axis is at $(0,3)$ .
Intersection at $(0,3)$ : The third vertex is the intersection of $x \geq 0$ and $y \geq 0$ , which is the origin $(0,0)$ .

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