Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+8 x-16$ $\newline$ 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}+8 x-16

\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 transform the given quadratic equation into vertex form. $\newline$ The given quadratic equation is $y = x^2 + 8x - 16$ . $\newline$ 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 $x^2 + 8x$ part. $\newline$ The value needed is $(8/2)^2 = 4^2 = 16$ .
Add and Subtract Value: Add and subtract the value found inside the equation. $\newline$ $y = x^2 + 8x + 16 - 16 - 16$ $\newline$ We added and subtracted $16$ to create a perfect square trinomial while keeping the equation balanced.
Rewrite with Perfect Square: Rewrite the equation with the perfect square trinomial and the constant term. $\newline$ $y = (x^2 + 8x + 16) - 32$ $\newline$ Now, the expression in the parentheses is a perfect square trinomial.
Factor Perfect Square: Factor the perfect square trinomial. $\newline$ $y = (x + 4)^2 - 32$ $\newline$ The factored form of $x^2 + 8x + 16$ is $(x + 4)^2$ .
Write in Vertex Form: Write the equation in vertex form and state the coordinates of the vertex. $\newline$ The vertex form of the equation is $y = (x + 4)^2 - 32$ . $\newline$ The vertex $(h, k)$ is found by considering the form $y = a(x - h)^2 + k$ . Here, $h = -4$ and $k = -32$ . $\newline$ So, the coordinates of the vertex are $(-4, -32)$ .