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

y=x^(2)-2x-35
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}-2 x-35$ $\newline$ 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}-2 x-35

\newline

Vertex Form:

y=

\newline

Vertex:

(\square, \square)

Identify Vertex Form: Identify the vertex form of a parabola. The vertex form of a parabola is given by $y = a(x - h)^2 + k$ , 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 $y = x^2 - 2x - 35$ . $\newline$ To complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial.
Calculate Value: Calculate the value needed to complete the square. $\newline$ The coefficient of $x$ is $-2$ . Half of $-2$ is $-1$ , and the square of $-1$ is $1$ . We will add and subtract $1$ inside the parentheses to complete the square.
Rewrite Equation: Rewrite the equation by adding and subtracting the calculated value. $\newline$ $y = x^2 - 2x + 1 - 1 - 35$ $\newline$ Now, group the perfect square trinomial and the constants. $\newline$ $y = (x^2 - 2x + 1) - 36$
Factor Trinomial: Factor the perfect square trinomial. $\newline$ $y = (x - 1)^2 - 36$ $\newline$ Now, the equation is in vertex form.
Identify Vertex: Identify the vertex of the parabola. $\newline$ The vertex form of the equation is $y = (x - 1)^2 - 36$ . $\newline$ Comparing this with the standard vertex form $y = a(x - h)^2 + k$ , we find that $h = 1$ and $k = -36$ . $\newline$ Therefore, the vertex of the parabola is $(1, -36)$ .

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