The equation of a parabola is y = x^2 - 6x + 16. 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 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. The given equation is y = x^2 - 6x + 16. To complete the square, we need to find the value that makes x^2 - 6x a perfect square trinomial. This value is (-6/2)^2 = 9. We will add and subtract 9 to the equation.Add and subtract: Add and subtract 9 to the equation. y = x^2 - 6x + 16 y = x^2 - 6x + 9 + 16 - 9 Now, we have the perfect square trinomial x^2 - 6x + 9 and the constants 16 - 9.Factor and simplify: Factor the perfect square trinomial and simplify the constants. y = (x^2 - 6x + 9) + 16 - 9 y = (x - 3)^2 + 7 Now, the equation is in vertex form, where (x - 3)^2 is the perfect square trinomial and +7 is the constant term.

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The equation of a parabola is $y = x^2 - 6x + 16$ . 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 + 16

. 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. $\newline$ The given equation is $y = x^2 - 6x + 16$ . To complete the square, we need to find the value that makes $x^2 - 6x$ a perfect square trinomial. This value is $(-6/2)^2 = 9$ . We will add and subtract $9$ to the equation.
Add and subtract: Add and subtract $9$ to the equation. $\newline$ $y = x^2 - 6x + 16$ $\newline$ $y = x^2 - 6x + 9 + 16 - 9$ $\newline$ Now, we have the perfect square trinomial $x^2 - 6x + 9$ and the constants $16 - 9$ .
Factor and simplify: Factor the perfect square trinomial and simplify the constants. $\newline$ $y = (x^2 - 6x + 9) + 16 - 9$ $\newline$ $y = (x - 3)^2 + 7$ $\newline$ Now, the equation is in vertex form, where $(x - 3)^2$ is the perfect square trinomial and $+7$ is the constant term.