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

y=x^(2)+10 x-2
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+10 x-2$ $\newline$ 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}+10 x-2

\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 given by $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Convert quadratic equation: Consider the given quadratic equation $y = x^2 + 10x - 2$ . $\newline$ To convert this into vertex form, we need to complete the square.
Factor out coefficient: Factor out the coefficient of the $x^2$ term if it is not $1$ . In this case, the coefficient of $x^2$ is $1$ , so we do not need to factor anything out.
Find square of half: Find the square of half the coefficient of the $x$ term to complete the square. Half of the coefficient of $x$ is $\frac{10}{2}$ , which is $5$ . Squaring $5$ gives us $25$ .
Add and subtract square: Add and subtract the square of half the coefficient of $x$ inside the equation. $\newline$ $y = x^2 + 10x + 25 - 25 - 2$ $\newline$ This allows us to form a perfect square trinomial while keeping the equation balanced.
Rewrite equation grouping: Rewrite the equation grouping the perfect square trinomial and combining the constants. $\newline$ $y = (x^2 + 10x + 25) - 27$
Factor perfect square trinomial: Factor the perfect square trinomial. $y = (x + 5)^2 - 27$
Write in vertex form: Write the equation in vertex form. $\newline$ Vertex Form: $y = (x + 5)^2 - 27$
Identify vertex coordinates: Identify the coordinates of the vertex. $\newline$ The vertex form of the equation is $y = a(x - h)^2 + k$ . Comparing this with our equation $y = (x + 5)^2 - 27$ , we find that $h = -5$ and $k = -27$ . $\newline$ Therefore, the coordinates of the vertex are $(-5, -27)$ .

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In which direction does the parabola

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The equation of a circle is

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. What are the center and radius of the circle?

\newline

1

\newline

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(-4,0)

and the radius is

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\newline

(B) The center is

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y=(x-3)(x+9)

\newline

The given equation is graphed in the

x y

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x

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\newline

1

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\newline

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Question

y=2 x^{2}-5 x+7

is graphed in the

x y

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\newline

1

\newline

x

\newline

y

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x

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\newline

y

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