Put the quadratic into vertex form and state the coordinates of the vertex. y=x^(2)+10 x+16 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. Given equation: y = x^2 + 10x + 16 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 10, so we take half of it, which is 5, and then square it to get 25.Add and Subtract: Add and subtract the square of half the coefficient of x inside the equation. y = x^2 + 10x + 25 - 25 + 16 We added and subtracted 25 to complete the square, and then we can group the perfect square trinomial and the constants.Rewrite Equation: Rewrite the equation with the perfect square trinomial and combine the constants. y = (x^2 + 10x + 25) - 9 Now, we can write the perfect square trinomial as a squared binomial. y = (x + 5)^2 - 9 This is the vertex form of the equation.Identify Vertex: Identify the vertex of the parabola. The vertex form of the equation is y = (x + 5)^2 - 9. The vertex (h, k) can be read directly from the vertex form as (h, k) = (-5, -9).

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

y=x^(2)+10 x+16
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+10 x+16$ $\newline$ 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}+10 x+16

\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. $\newline$ Given equation: $y = x^2 + 10x + 16$ $\newline$ 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 $10$ , so we take half of it, which is $5$ , and then square it to get $25$ .
Add and Subtract: Add and subtract the square of half the coefficient of $x$ inside the equation. $\newline$ $y = x^2 + 10x + 25 - 25 + 16$ $\newline$ We added and subtracted $25$ to complete the square, and then we can group the perfect square trinomial and the constants.
Rewrite Equation: Rewrite the equation with the perfect square trinomial and combine the constants. $\newline$ $y = (x^2 + 10x + 25) - 9$ $\newline$ Now, we can write the perfect square trinomial as a squared binomial. $\newline$ $y = (x + 5)^2 - 9$ $\newline$ This is the vertex form of the equation.
Identify Vertex: Identify the vertex of the parabola. $\newline$ The vertex form of the equation is $y = (x + 5)^2 - 9$ . $\newline$ The vertex $(h, k)$ can be read directly from the vertex form as $(h, k) = (-5, -9)$ .