Put the quadratic into vertex form and state the coordinates of the vertex. y=x^(2)+10 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. Given equation: y = x^2 + 10x + 24 To complete the square, we need to find a number that, when added and subtracted to the equation, forms a perfect square trinomial with x^2 + 10x. Half of the linear coefficient (10) is 5, and squaring it gives us 25. So, we add and subtract 25 inside the equation to complete the square. y = x^2 + 10x + 25 - 25 + 24Rewrite equation and combine: Rewrite the equation with the perfect square trinomial and combine the constants. y = (x^2 + 10x + 25) - 1 Now, factor the perfect square trinomial. y = (x + 5)^2 - 1 This is the vertex form of the equation.Identify vertex: Identify the vertex of the parabola. From the vertex form y = (x + 5)^2 - 1, we can see that the vertex (h, k) is (-5, -1).

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

y=x^(2)+10 x+24
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}+10 x+24$ $\newline$ 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}+10 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. $\newline$ Given equation: $y = x^2 + 10x + 24$ $\newline$ To complete the square, we need to find a number that, when added and subtracted to the equation, forms a perfect square trinomial with $x^2 + 10x$ . $\newline$ Half of the linear coefficient ( $10$ ) is $5$ , and squaring it gives us $25$ . $\newline$ So, we add and subtract $25$ inside the equation to complete the square. $\newline$ $y = x^2 + 10x + 25 - 25 + 24$
Rewrite equation and combine: Rewrite the equation with the perfect square trinomial and combine the constants. $\newline$ $y = (x^2 + 10x + 25) - 1$ $\newline$ Now, factor the perfect square trinomial. $\newline$ $y = (x + 5)^2 - 1$ $\newline$ This is the vertex form of the equation.
Identify vertex: Identify the vertex of the parabola. $\newline$ From the vertex form $y = (x + 5)^2 - 1$ , we can see that the vertex $(h, k)$ is $(-5, -1)$ .