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

y=x^(2)-8x+41
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}-8 x+41$ $\newline$ 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}-8 x+41

\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: $y = x^2 - 8x + 41$ $\newline$ We need to find a value to add and subtract to complete the square. $\newline$ The coefficient of $x$ is $-8$ , so we take half of it, which is $-4$ , and then square it to get $16$ . $\newline$ We will add and subtract $16$ inside the equation to complete the square.
Add/subtract square: Add and subtract the square of half the coefficient of $x$ inside the equation. $\newline$ $y = x^2 - 8x + 16 - 16 + 41$ $\newline$ Now, group the perfect square trinomial and the constants. $\newline$ $y = (x^2 - 8x + 16) - 16 + 41$
Factor and simplify: Factor the perfect square trinomial and simplify the constants. $\newline$ $y = (x - 4)^2 - 16 + 41$ $\newline$ $y = (x - 4)^2 + 25$ $\newline$ Now we have the equation in vertex form.
Identify parabola vertex: Identify the vertex of the parabola. $\newline$ The vertex form of the equation is $y = (x - 4)^2 + 25$ . $\newline$ Comparing this with the standard vertex form $y = a(x - h)^2 + k$ , we find that $h = 4$ and $k = 25$ . $\newline$ Therefore, the vertex of the parabola is $(4, 25)$ .

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