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The equation of a parabola is $y = x^2 + 6x + 1$ . Write the equation in vertex form. $\newline$ Write any numbers as integers or simplified proper or improper fractions. $\newline$ ______

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Write a quadratic function in vertex form

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Q. The equation of a parabola is

y = x^2 + 6x + 1

. Write the equation in vertex form.

\newline

Write any numbers as integers or simplified proper or improper fractions.

\newline

______

Identify vertex form: Identify the vertex form of a parabola. $\newline$ The vertex form of a parabola is given by the equation $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Complete the square: Complete the square to rewrite the given equation in vertex form. $\newline$ We start with the given equation $y = x^2 + 6x + 1$ . To complete the square, we need to find the value that makes $x^2 + 6x$ a perfect square trinomial. We do this by taking half of the coefficient of $x$ , which is $6$ , dividing it by $2$ to get $3$ , and then squaring it to get $9$ . We will add and subtract this value inside the equation.
Add and subtract value: Add and subtract the value found in Step $2$ inside the equation. $\newline$ $y = x^2 + 6x + 9 - 9 + 1$ $\newline$ Now we have added $9$ and subtracted $9$ , which keeps the equation balanced.
Group and combine: Group the perfect square trinomial and combine the constants. $\newline$ $y = (x^2 + 6x + 9) - 9 + 1$ $\newline$ $y = (x + 3)^2 - 8$ $\newline$ Now we have the equation in vertex form, where $(x + 3)^2$ is the perfect square trinomial and $-8$ is the combination of the constants $-9 + 1$ .

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Solve by completing the square.

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