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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Consider the equation below.

y=3x^2+30x+71
y=3(x+__)^2+__

Use completing the square to rewrite the given equation and reveal the extreme value.

The extreme value of the equation is at (__ ,__ )



(the ___ are blanks where the answers are supposed to be)

Respuesta :

frika
If [tex]y=3x^2+30x+71[/tex], then 
[tex]y=3(x^2+10x)+71=3(x^2+2\cdot5x+25-25)+71[/tex]
[tex]y=3((x+5)^2-25)+71=3(x+5)^2-75+71=3(x+5)^2-4[/tex]. So, in the first gap you can write 5 and in the second you can write -4.
Since [tex](x+5)^2\ge 0[/tex], the minimum value of y is when [tex](x+5)^2=0[/tex]. The solution of the last equation is x=-5 and then y=-4 -- these are numbers, which you can write in the third and fourth gaps.

autumnmcdaniels

Answer:

the last equation is x=-5 and then y=-4 -- these are numbers, which you can write in the third and fourth gaps.

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