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Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an
(x,y) point.

y=-x^(2)-6
Answer:

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an $(x, y)$ point. $\newline$ $y=-x^{2}-6$ $\newline$ Answer:

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Solve exponential equations by rewriting the base

Full solution

Q. Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an

(x, y)

point.

\newline

y=-x^{2}-6

\newline

Answer:

Convert to Vertex Form: The vertex form of a parabola is given by $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. To find the vertex of the parabola $y = -x^2 - 6$ , we need to complete the square to convert the equation into vertex form.
Identify Vertex: Since the equation is already in the form $y = ax^2 + bx + c$ , with $a = -1$ , $b = 0$ , and $c = -6$ , there is no need to complete the square because the $x$ -term is missing ( $b = 0$ ). This means the vertex occurs at $x = 0$ .
Calculate y-coordinate: To find the y-coordinate of the vertex, we substitute $x = 0$ into the equation $y = -x^2 - 6$ . $\newline$ $y = -(0)^2 - 6$ $\newline$ $y = -6$
Find Vertex Coordinates: The coordinates of the vertex are $(0, -6)$ .

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