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

y=x^(2)-4x-12
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}-4 x-12$ $\newline$ 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}-4 x-12

\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: $y = x^2 - 4x - 12$ $\newline$ To complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial with the $x^2$ and $-4x$ terms. $\newline$ The value to complete the square is found by taking half of the coefficient of $x$ and squaring it: $(-4/2)^2 = 2^2 = 4$ .
Add and Subtract Value: Add and subtract the value found in Step $2$ inside the equation. $\newline$ $y = x^2 - 4x + 4 - 4 - 12$ $\newline$ $y = (x^2 - 4x + 4) - 16$ $\newline$ Now, the equation has a perfect square trinomial $x^2 - 4x + 4$ which can be written as $(x - 2)^2$ .
Rewrite in Vertex Form: Rewrite the equation in vertex form using the perfect square trinomial. $\newline$ $y = (x - 2)^2 - 16$ $\newline$ This is the vertex form of the given quadratic equation.
Identify Vertex Coordinates: Identify the coordinates of the vertex from the vertex form. $\newline$ The vertex form of the equation is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex. $\newline$ From the equation $y = (x - 2)^2 - 16$ , we can see that $h = 2$ and $k = -16$ . $\newline$ Therefore, the coordinates of the vertex are $(2, -16)$ .

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