The equation of a parabola is y = x^2 + 10x + 29. 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 + 10x + 29. To complete the square, we need to find the value that makes x^2 + 10x a perfect square trinomial. This value is (10/2)^2 = 25. We will add and subtract 25 inside the equation.Add and Subtract: Add and subtract 25 to the equation. y = x^2 + 10x + 29 y = x^2 + 10x + 25 + 29 - 25 We added 25 and subtracted 25, so the equation remains equivalent to the original.Rewrite and Simplify: Rewrite the equation with the perfect square trinomial and simplify. y = (x^2 + 10x + 25) + 29 - 25 y = (x + 5)^2 + 4 Now the equation is in vertex form, where (x + 5)^2 is the perfect square trinomial and 4 is the constant term.

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The equation of a parabola is $y = x^2 + 10x + 29$ . 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 + 10x + 29

. 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. 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 $y = x^2 + 10x + 29$ . To complete the square, we need to find the value that makes $x^2 + 10x$ a perfect square trinomial. This value is $(10/2)^2 = 25$ . We will add and subtract $25$ inside the equation.
Add and Subtract: Add and subtract $25$ to the equation. $\newline$ $y = x^2 + 10x + 29$ $\newline$ $y = x^2 + 10x + 25 + 29 - 25$ $\newline$ We added $25$ and subtracted $25$ , so the equation remains equivalent to the original.
Rewrite and Simplify: Rewrite the equation with the perfect square trinomial and simplify. $\newline$ $y = (x^2 + 10x + 25) + 29 - 25$ $\newline$ $y = (x + 5)^2 + 4$ $\newline$ Now the equation is in vertex form, where $(x + 5)^2$ is the perfect square trinomial and $4$ is the constant term.