Respuesta :

calculista

Answer:

[tex]y=6(x+1)^{2}-16[/tex]

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

[tex]y=a(x-h)^{2}+k[/tex]

where

(h,k) is the vertex of the parabola

In this problem we have

[tex]y=6x^{2}+12x-10[/tex]

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

[tex]y+10=6x^{2}+12x[/tex]

Factor the leading coefficient

[tex]y+10=6(x^{2}+2x)[/tex]

Complete the square. Remember to balance the equation by adding the same constants to each side

[tex]y+10+6=6(x^{2}+2x+1)[/tex]

[tex]y+16=6(x^{2}+2x+1)[/tex]

Rewrite as perfect squares

[tex]y+16=6(x+1)^{2}[/tex]

[tex]y=6(x+1)^{2}-16[/tex] -------> equation in vertex form

The vertex is the point [tex](-1,-16)[/tex]


adioabiola

The equation y = 6x2 + 12x – 10 can be written in vertex form as; y = 6(x + 1)² - 16.

To re-write the equation y = 6x² + 12x – 10 in vertex form, the completing the square method is used as follows;

  • y = 6x² + 12x – 10

First, we isolate x- terms and factorise;

  • y + 10 = 6(x² + 2x)

  • y + 10 + 6 = 6 (x² + 2x + 1)

By rewriting as perfect squares: we have

  • y + 16 = 6(x + 1)²

Consequently, the equation in vertex form is;

  • y = 6(x + 1)² - 16

The vertex is therefore given by coordinates; (-1,-16).

Read more:

https://brainly.com/question/11376583

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