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Put the quadratic into vertex form and state the coordinates of the vertex.

y=x^(2)+4x+29
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+4 x+29$ $\newline$ Vertex Form: $y=$ $\newline$ Vertex: $(\square, \square)$

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Convert equations of parabolas from general to vertex form

Full solution

Q. Put the quadratic into vertex form and state the coordinates of the vertex.

\newline

y=x^{2}+4 x+29

\newline

Vertex Form:

y=

\newline

Vertex:

(\square, \square)

Identify vertex form: Identify the vertex form of a parabola. $\newline$ The vertex form of a parabola is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Complete the square: Complete the square to rewrite the quadratic equation in vertex form. $\newline$ Given equation: $y = x^2 + 4x + 29$ $\newline$ We need to find a value to add and subtract to complete the square. $\newline$ The coefficient of $x$ is $4$ , so we take half of it, which is $2$ , and then square it to get $4$ . $\newline$ We add and subtract $4$ inside the parentheses to complete the square.
Rewrite equation: Rewrite the equation by completing the square. $\newline$ $y = x^2 + 4x + 4 - 4 + 29$ $\newline$ $y = (x^2 + 4x + 4) - 4 + 29$ $\newline$ $y = (x + 2)^2 + 25$ $\newline$ Now the equation is in vertex form.
Identify vertex: Identify the vertex of the parabola. The vertex form of the equation is $y = (x + 2)^2 + 25$ . Comparing this with the standard vertex form $y = a(x - h)^2 + k$ , we find that $h = -2$ and $k = 25$ . Therefore, the vertex of the parabola is $(-2, 25)$ .

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Question

The equation of a circle is

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\newline

1

\newline

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and the radius is

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\newline

(B) The center is

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\newline

The given equation is graphed in the

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\newline

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x

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y

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\newline

y

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