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

y=x^(2)-16 x
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}-16 x$ $\newline$ 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}-16 x

\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 given equation $y = x^2 - 16x$ in vertex form. $\newline$ First, factor out the coefficient of the $x^2$ term, which is $1$ in this case, so we can leave it as is. $\newline$ Next, take half of the coefficient of the $x$ term, which is $-16$ , and square it to complete the square. $\newline$ Half of $-16$ is $-8$ , and $(-8)^2 = 64$ .
Add and subtract terms: Add and subtract the square of half the coefficient of $x$ inside the equation. $\newline$ $y = x^2 - 16x + 64 - 64$ $\newline$ We added $64$ to complete the square and then subtracted $64$ to keep the equation balanced.
Rewrite equation by grouping: Rewrite the equation by grouping the perfect square trinomial and the constant term. $\newline$ $y = (x^2 - 16x + 64) - 64$ $\newline$ $y = (x - 8)^2 - 64$ $\newline$ Now the equation is in vertex form.
Identify vertex: Identify the vertex of the parabola. $\newline$ The vertex form of the equation is $y = (x - 8)^2 - 64$ . $\newline$ Comparing this with the standard vertex form $y = a(x - h)^2 + k$ , we find that $h = 8$ and $k = -64$ . $\newline$ Therefore, the vertex of the parabola is $(8, -64)$ .

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