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

y=x^(2)-12 x+32
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}-12 x+32$ $\newline$ 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}-12 x+32

\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: $y = x^2 - 12x + 32$ $\newline$ To complete the square, we need to find the value that makes $x^2 - 12x$ a perfect square trinomial. $\newline$ We take half of the coefficient of $x$ , which is $-12/2 = -6$ , and square it to get $36$ .
Add and Subtract: Add and subtract the square of half the coefficient of $x$ inside the equation. $\newline$ $y = x^2 - 12x + 36 - 36 + 32$ $\newline$ We added and subtracted $36$ to complete the square while keeping the equation balanced.
Rewrite Equation: Rewrite the equation by grouping the perfect square trinomial and combining the constants. $\newline$ $y = (x^2 - 12x + 36) - 4$ $\newline$ $y = (x - 6)^2 - 4$ $\newline$ Now the equation is in vertex form.
Identify Vertex: Identify the vertex of the parabola. $\newline$ The vertex form of the equation is $y = (x - 6)^2 - 4$ . $\newline$ Comparing this with the standard vertex form $y = a(x - h)^2 + k$ , we find that $h = 6$ and $k = -4$ . $\newline$ Therefore, the vertex of the parabola is $(6, -4)$ .

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