Put the quadratic into vertex form and state the coordinates of the vertex. y=x^(2)+2x+37 Vertex Form: y= Vertex: (◻,◻)

Identify Vertex Form: Identify the vertex form of a parabola. 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 y = x^2 + 2x + 37 in vertex form. First, we need to create a perfect square trinomial from the quadratic and linear terms. To do this, we take half of the linear coefficient (2), square it, and add and subtract this value inside the equation. Half of the linear coefficient: 2/2 = 1 Square of half the linear coefficient: 1^2 = 1Add and Subtract: Add and subtract the square of half the linear coefficient inside the equation. y = x^2 + 2x + 1 - 1 + 37 Now, group the perfect square trinomial and combine the constants. y = (x^2 + 2x + 1) + (37 - 1) y = (x + 1)^2 + 36Write in Vertex Form: Write the equation in vertex form and identify the vertex. The vertex form of the equation is y = (x + 1)^2 + 36. The vertex (h, k) can be read directly from the vertex form as (-1, 36).

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

y=x^(2)+2x+37
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+2 x+37$ $\newline$ Vertex Form: $y=$ $\newline$ Vertex: $(\square, \square)$

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Algebra 2

Convert equations of parabolas from general to vertex form

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

\newline

y=x^{2}+2 x+37

\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 $y = x^2 + 2x + 37$ in vertex form. $\newline$ First, we need to create a perfect square trinomial from the quadratic and linear terms. To do this, we take half of the linear coefficient ( $2$ ), square it, and add and subtract this value inside the equation. $\newline$ Half of the linear coefficient: $\frac{2}{2} = 1$ $\newline$ Square of half the linear coefficient: $1^2 = 1$
Add and Subtract: Add and subtract the square of half the linear coefficient inside the equation. $\newline$ $y = x^2 + 2x + 1 - 1 + 37$ $\newline$ Now, group the perfect square trinomial and combine the constants. $\newline$ $y = (x^2 + 2x + 1) + (37 - 1)$ $\newline$ $y = (x + 1)^2 + 36$
Write in Vertex Form: Write the equation in vertex form and identify the vertex. $\newline$ The vertex form of the equation is $y = (x + 1)^2 + 36$ . $\newline$ The vertex $(h, k)$ can be read directly from the vertex form as $(-1, 36)$ .