Put the quadratic into vertex form and state the coordinates of the vertex. y=x^(2)+14 x+24 Vertex Form: y= Vertex: (◻,◻)

Identify vertex form: Identify the vertex form of a parabola. The vertex form of a parabola is y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.Complete the square: Complete the square to rewrite the quadratic equation in vertex form. The given equation is y = x^2 + 14x + 24. To complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial. The coefficient of x is 14, so we take half of it, which is 7, and then square it to get 49. We add and subtract 49 inside the equation to complete the square.Add and subtract: Add and subtract 49 inside the equation. y = x^2 + 14x + 49 - 49 + 24 Now, group the perfect square trinomial and the constants. y = (x^2 + 14x + 49) - 25Factor perfect square: Factor the perfect square trinomial. The perfect square trinomial x^2 + 14x + 49 can be factored into (x + 7)^2. So the equation becomes: y = (x + 7)^2 - 25Write in vertex form: Write the equation in vertex form and state the coordinates of the vertex. The vertex form of the equation is y = (x + 7)^2 - 25. The vertex (h, k) is the point where the equation is in the form (x - h)^2 + k. In our equation, h is -7 and k is -25, so the vertex is (-7, -25).

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Put the quadratic into vertex form and state the coordinates of the vertex.

y=x^(2)+14 x+24
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+14 x+24$ $\newline$ Vertex Form: $y=$ $\newline$ Vertex: $(\square, \square)$

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Algebra 2

Convert equations of parabolas from general to vertex form

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Q. Put the quadratic into vertex form and state the coordinates of the vertex.

\newline

y=x^{2}+14 x+24

\newline

Vertex Form:

y=

\newline

Vertex:

(\square, \square)

Identify vertex form: Identify the vertex form of a parabola. $\newline$ The vertex form of a parabola is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.
Complete the square: Complete the square to rewrite the quadratic equation in vertex form. $\newline$ The given equation is $y = x^2 + 14x + 24$ . $\newline$ To complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial. $\newline$ The coefficient of $x$ is $14$ , so we take half of it, which is $7$ , and then square it to get $49$ . $\newline$ We add and subtract $49$ inside the equation to complete the square.
Add and subtract: Add and subtract $49$ inside the equation. $\newline$ $y = x^2 + 14x + 49 - 49 + 24$ $\newline$ Now, group the perfect square trinomial and the constants. $\newline$ $y = (x^2 + 14x + 49) - 25$
Factor perfect square: Factor the perfect square trinomial. $\newline$ The perfect square trinomial $x^2 + 14x + 49$ can be factored into $(x + 7)^2$ . $\newline$ So the equation becomes: $\newline$ $y = (x + 7)^2 - 25$
Write in vertex form: Write the equation in vertex form and state the coordinates of the vertex. $\newline$ The vertex form of the equation is $y = (x + 7)^2 - 25$ . $\newline$ The vertex $(h, k)$ is the point where the equation is in the form $(x - h)^2 + k$ . $\newline$ In our equation, $h$ is $-7$ and $k$ is $-25$ , so the vertex is $(-7, -25)$ .