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

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 equation into vertex form. We have the equation y = x^2 + 6x - 1. To complete the square, we need to find a number that, when added and subtracted to the equation, forms a perfect square trinomial with x^2 + 6x. This number is (6/2)^2 = 3^2 = 9. We add and subtract 9 within the equation.Add and subtract: Add and subtract 9 to the equation. y = x^2 + 6x - 1 y = x^2 + 6x + 9 - 9 - 1 Now, we have x^2 + 6x + 9 as a perfect square trinomial.Factor and simplify: Factor the perfect square trinomial and simplify the equation. y = (x^2 + 6x + 9) - 9 - 1 y = (x + 3)^2 - 10 Now, the equation is in vertex form, where (x + 3)^2 is the perfect square trinomial and -10 is the constant term.

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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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Algebra 1

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 $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Complete the square: Complete the square to transform the given equation into vertex form. $\newline$ We have the equation $y = x^2 + 6x - 1$ . To complete the square, we need to find a number that, when added and subtracted to the equation, forms a perfect square trinomial with $x^2 + 6x$ . This number is $(6/2)^2 = 3^2 = 9$ . We add and subtract $9$ within the equation.
Add and subtract: Add and subtract $9$ to the equation. $\newline$ $y = x^2 + 6x - 1$ $\newline$ $y = x^2 + 6x + 9 - 9 - 1$ $\newline$ Now, we have $x^2 + 6x + 9$ as a perfect square trinomial.
Factor and simplify: Factor the perfect square trinomial and simplify the equation. $\newline$ $y = (x^2 + 6x + 9) - 9 - 1$ $\newline$ $y = (x + 3)^2 - 10$ $\newline$ Now, the equation is in vertex form, where $(x + 3)^2$ is the perfect square trinomial and $-10$ is the constant term.