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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)-16 x-57
Answer:

Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an $(x, y)$ point. $\newline$ $y=-x^{2}-16 x-57$ $\newline$ Answer:

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Find the axis of symmetry of a parabola

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}-16 x-57

\newline

Answer:

Factor out x terms: The vertex form of a parabola's equation is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola. To find the vertex of the parabola given by $y = -x^2 - 16x - 57$ , we need to complete the square to rewrite the equation in vertex form.
Complete the square: First, we factor out the coefficient of the $x^2$ term from the $x$ terms. Since the coefficient is $-1$ , we have: $\newline$ $y = -1(x^2 + 16x) - 57$
Rewrite the equation: Next, we find the value that completes the square for the expression $x^2 + 16x$ . This value is $(16/2)^2 = 64$ . We add and subtract this value inside the parentheses to complete the square: $\newline$ $y = -1(x^2 + 16x + 64 - 64) - 57$
Distribute and simplify: Now we rewrite the equation, grouping the terms that create a perfect square trinomial and combining the constants: $y = -1((x + 8)^2 - 64) - 57$
Final vertex form: Distribute the $-1$ and simplify the constants: $\newline$ $y = -(x + 8)^2 + 64 - 57$ $\newline$ $y = -(x + 8)^2 + 7$
Final vertex form: Distribute the $-1$ and simplify the constants: $\newline$ $y = -(x + 8)^2 + 64 - 57$ $\newline$ $y = -(x + 8)^2 + 7$ Now the equation is in vertex form, $y = a(x - h)^2 + k$ , where $a = -1$ , $h = -8$ , and $k = 7$ . Therefore, the vertex of the parabola is $(h, k) = (-8, 7)$ .

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