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

y=x^(2)+2x+3
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+2 x+3$ $\newline$ 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}+2 x+3

\newline

Vertex Form:

y=

\newline

Vertex:

(\square, \square)

Identify Vertex Form: Identify the vertex form of a parabola. $\newline$ Vertex form: $y = a(x - h)^2 + k$
Complete the Square: Complete the square for the quadratic equation $y = x^2 + 2x + 3$ . $\newline$ To complete the square, we need to find a value that makes $x^2 + 2x + \_\_\_$ a perfect square trinomial. $\newline$ The value needed is $(2/2)^2 = 1^2 = 1$ .
Rewrite Equation: Rewrite the equation by adding and subtracting the value found in Step $2$ inside the parentheses. $\newline$ $y = x^2 + 2x + 1 - 1 + 3$ $\newline$ $y = (x^2 + 2x + 1) + 2$ $\newline$ Now, the equation inside the parentheses is a perfect square trinomial.
Factor and Simplify: Factor the perfect square trinomial and simplify the equation. $\newline$ $y = (x + 1)^2 + 2$ $\newline$ This is the vertex form of the quadratic equation.
Identify Vertex: Identify the vertex of the parabola from the vertex form. $\newline$ The vertex form $y = a(x - h)^2 + k$ gives the vertex as the point $(h, k)$ . $\newline$ From $y = (x + 1)^2 + 2$ , we can see that $h = -1$ and $k = 2$ . $\newline$ Therefore, the vertex is $(-1, 2)$ .

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Question

The equation of a circle is

x^{2}+(y+4)^{2}=1

. What are the center and radius of the circle?

\newline

1

\newline

(A) The center is

(-4,0)

and the radius is

1

\newline

(B) The center is

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and the radius is

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(D) The center is

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Question

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

\newline

The given equation is graphed in the

x y

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\newline

1

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Question

y=2 x^{2}-5 x+7

is graphed in the

x y

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\newline

1

\newline

x

\newline

y

\newline

x

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\newline

y

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