Respuesta :

kudzordzifrancis
ANSWER
[tex]y = 3{(x + 5})^{2}-4[/tex]

The extreme values are

(-5,-4)

EXPLANATION

The given equation is

[tex]y = 3 {x}^{2} + 30x+71[/tex]

Factor 3 from the first two terms,

[tex]y = 3( {x}^{2} + 10x) + 71[/tex]

Add and subtract the square of half the coefficient of x.

[tex]y = 3( {x}^{2} + 10x + 25) + 71 - 3(25)[/tex]

[tex]y = 3{(x + 5})^{2}+ 71-75[/tex]

[tex]y = 3{(x + 5})^{2} -4[/tex]

The extreme values are

(-5,-4)
alinakincsem

Answer:

[tex]y= 3(x+5)^2 -4[/tex], (-5, -4)

Step-by-step explanation:

We are given the following equation and we are to complete its square:

[tex] y = 3x^2 + 30x + 71 [/tex]

[tex] y = 3(x^2 + 10x) + 71 [/tex]

[tex] y = 3(x^2 + 10x + 5^2 - 5^2) + 71 [/tex]

[tex]y= 3(x^2 +10x +25) +3(-25) + 71[/tex]

[tex]y= 3(x+5)(x+5) -75 + 71[/tex]

[tex]y= 3(x+5)^2 -4[/tex]

So out extreme values of the equation is (-5, -4).

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