The equation of a parabola is y = x^2 + 6x + 11. 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 square transformation: Complete the square to transform the given equation into vertex form. We have the equation y = x^2 + 6x + 11. 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 = 3^2 = 9. We will add and subtract 9 to the equation.Add and subtract value: Add and subtract the value found in Step 2 to the equation. y = x^2 + 6x + 9 - 9 + 11 Now, group the perfect square trinomial and the constants separately. y = (x^2 + 6x + 9) - 9 + 11Rewrite perfect square trinomial: Rewrite the perfect square trinomial as a squared binomial. The perfect square trinomial x^2 + 6x + 9 can be factored into (x + 3)^2. Now, combine the constants -9 + 11 to get 2. y = (x + 3)^2 + 2Write final equation: Write the final equation in vertex form. The equation of the parabola in vertex form is y = (x + 3)^2 + 2.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The equation of a parabola is $y = x^2 + 6x + 11$ . Write the equation in vertex form. $\newline$ Write any numbers as integers or simplified proper or improper fractions. $\newline$ ______

View step-by-step help

Home

Math Problems

Algebra 1

Write a quadratic function in vertex form

Full solution

Q. The equation of a parabola is

y = x^2 + 6x + 11

. 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 square transformation: Complete the square to transform the given equation into vertex form. $\newline$ We have the equation $y = x^2 + 6x + 11$ . To complete the square, we need to find the value that makes $x^2 + 6x$ a perfect square trinomial. This value is $(\frac{6}{2})^2 = 3^2 = 9$ . We will add and subtract $9$ to the equation.
Add and subtract value: Add and subtract the value found in Step $2$ to the equation. $\newline$ $y = x^2 + 6x + 9 - 9 + 11$ $\newline$ Now, group the perfect square trinomial and the constants separately. $\newline$ $y = (x^2 + 6x + 9) - 9 + 11$
Rewrite perfect square trinomial: Rewrite the perfect square trinomial as a squared binomial. $\newline$ The perfect square trinomial $x^2 + 6x + 9$ can be factored into $(x + 3)^2$ . Now, combine the constants $-9 + 11$ to get $2$ . $\newline$ $y = (x + 3)^2 + 2$
Write final equation: Write the final equation in vertex form. $\newline$ The equation of the parabola in vertex form is $y = (x + 3)^2 + 2$ .