Put the quadratic into vertex form and state the coordinates of the vertex. y=x^(2)-4x-32 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 - 4x - 32 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 -4, so we take half of it, which is -2, and then square it to get 4. We add and subtract 4 inside the equation to complete the square.Add and subtract values: Add and subtract the value found inside the parentheses. y = x^2 - 4x + 4 - 4 - 32 Now, group the perfect square trinomial and the constants. y = (x^2 - 4x + 4) - 36Factor perfect square trinomial: Factor the perfect square trinomial. y = (x - 2)^2 - 36 Now we have the equation in vertex form.Identify vertex: Identify the vertex of the parabola. The vertex form of the equation is y = (x - 2)^2 - 36. Comparing this with the standard vertex form y = a(x - h)^2 + k, we find that h = 2 and k = -36. Therefore, the vertex of the parabola is (2, -36).

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

y=x^(2)-4x-32
Vertex Form:
y=
Vertex:
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex. $\newline$ $y=x^{2}-4 x-32$ $\newline$ 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}-4 x-32

\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: $y = x^2 - 4x - 32$ $\newline$ 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 $-4$ , so we take half of it, which is $-2$ , and then square it to get $4$ . $\newline$ We add and subtract $4$ inside the equation to complete the square.
Add and subtract values: Add and subtract the value found inside the parentheses. $\newline$ $y = x^2 - 4x + 4 - 4 - 32$ $\newline$ Now, group the perfect square trinomial and the constants. $\newline$ $y = (x^2 - 4x + 4) - 36$
Factor perfect square trinomial: Factor the perfect square trinomial. $\newline$ $y = (x - 2)^2 - 36$ $\newline$ Now we have the equation in vertex form.
Identify vertex: Identify the vertex of the parabola. $\newline$ The vertex form of the equation is $y = (x - 2)^2 - 36$ . $\newline$ Comparing this with the standard vertex form $y = a(x - h)^2 + k$ , we find that $h = 2$ and $k = -36$ . $\newline$ Therefore, the vertex of the parabola is $(2, -36)$ .