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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:
Submit Answer

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: $\newline$ Submit Answer

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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+37

\newline

Vertex Form:

y=

\newline

Vertex:

\newline

Submit Answer

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 rewrite the quadratic equation $y = x^2 + 2x + 37$ in vertex form. $\newline$ First, we need to focus on the $x$ -terms. We have $x^2 + 2x$ . To complete the square, we take half of the coefficient of $x$ , which is $\frac{2}{2} = 1$ , and square it, giving us $1^2 = 1$ . We will add and subtract this value inside the parentheses to maintain equality.
Rewrite Equation: Rewrite the equation by adding and subtracting the squared term. $\newline$ $y = x^2 + 2x + 1 - 1 + 37$ $\newline$ Now, group the perfect square trinomial and the constants. $\newline$ $y = (x^2 + 2x + 1) + 36$
Factor Trinomial: Factor the perfect square trinomial. $\newline$ The expression $x^2 + 2x + 1$ factors to $(x + 1)^2$ . $\newline$ So, the equation now reads $y = (x + 1)^2 + 36$ .
Write in Vertex Form: Write the equation in vertex form and identify the vertex. $\newline$ The equation in vertex form is $y = (x + 1)^2 + 36$ . $\newline$ The vertex of the parabola, given by the form $y = a(x - h)^2 + k$ , is $(h, k)$ . In our equation, $h = -1$ and $k = 36$ . $\newline$ Therefore, the vertex is $(-1, 36)$ .

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