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Look at the system of inequalities. $\newline$ $y \leq -x + 9$ $\newline$ $x \geq 0$ $\newline$ $y \geq 0$ $\newline$ 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 -x + 9

\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 $y \leq -x + 9$ and $x \geq 0$ . Set $x = 0$ in the first inequality: $y \leq 9$ . So one vertex is at $(0, 9)$ .
Identify Vertex at $(0, 9)$ : Next, find the intersection of $y \leq -x + 9$ and $y \geq 0$ . Set $y = 0$ in the first inequality: $0 \leq -x + 9$ . Solve for $x$ : $x \leq 9$ . So another vertex is at $(9, 0)$ .
Identify Vertex at $(9, 0)$ : Finally, find the intersection of $x \geq 0$ and $y \geq 0$ . This is simply the origin, so the last vertex is at $(0, 0)$ .

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Question

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\newline

1

\newline

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\newline

(B) The center is

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y=(x-3)(x+9)

\newline

The given equation is graphed in the

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1

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-3

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1

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y

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