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

y=x^(2)+14 x+22
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+14 x+22$ $\newline$ 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}+14 x+22

\newline

Vertex Form:

y=

\newline

Vertex:

(\square, \square)

Identify Vertex Form: Identify the vertex form of a parabola. The vertex form of a parabola is given by $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Complete the Square: Complete the square to transform the given quadratic equation $y = x^2 + 14x + 22$ into vertex form. $\newline$ First, we need to factor out the coefficient of the $x^2$ term if it is not $1$ . In this case, it is already $1$ , so we can proceed to the next step.
Find Half Coefficient Square: Find the square of half the coefficient of the $x$ term to complete the square. $\newline$ The coefficient of the $x$ term is $14$ , so half of it is $7$ . Squaring $7$ gives us $49$ .
Add/Subtract to Complete Square: Add and subtract the square of half the coefficient of $x$ inside the equation. $\newline$ $y = x^2 + 14x + 49 - 49 + 22$ $\newline$ This allows us to form a perfect square trinomial while keeping the equation balanced.
Rewrite Equation in Vertex Form: Rewrite the equation grouping the perfect square trinomial and combining the constants. $\newline$ $y = (x^2 + 14x + 49) - 27$ $\newline$ $y = (x + 7)^2 - 27$ $\newline$ Now the equation is in vertex form.
Identify Parabola Vertex: Identify the vertex of the parabola using the vertex form. The vertex form is $y = (x + 7)^2 - 27$ , so the vertex $(h, k)$ is $(-7, -27)$ .

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\newline

1

\newline

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(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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x

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y

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x

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\newline

y

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