Question

Consider the equation below. y=3x^2+30x+71

Answer 1

`y= 3(x+5)^2 -4`, (-5, -4)

Explanation:

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

`y = 3x^2 + 30x + 71`

`y = 3(x^2 + 10x) + 71`

`y = 3(x^2 + 10x + 5^2 - 5^2) + 71`

`y= 3(x^2 +10x +25) +3(-25) + 71`

`y= 3(x+5)(x+5) -75 + 71`

`y= 3(x+5)^2 -4`

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

Answer 2

ANSWER

`y = 3{(x + 5})^(2)-4`

The extreme values are

(-5,-4)

EXPLANATION

The given equation is


`y = 3 {x}^(2) + 30x+71`

Factor 3 from the first two terms,


`y = 3( {x}^(2) + 10x) + 71`

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


`y = 3( {x}^(2) + 10x + 25) + 71 - 3(25)`


`y = 3{(x + 5})^(2)+ 71-75`


`y = 3{(x + 5})^(2) -4`

The extreme values are

(-5,-4)