Answer:
The extreme value of the equation is (-5,-4)
Step-by-step explanation:
The complete question is shown in the attachment.
The given equation is [tex]y=3x^2+30x+71[/tex]
Let us factor 3 out of 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+(5)^2)-3( 5)^2+71[/tex]
We simplify (observe that the first three terms form a perfect square trinomial)
[tex]y=3(x+5)^2-3*25+71[/tex]
[tex]y=3(x+5)^2-75+71[/tex]
[tex]y=3(x+5)^2-4[/tex]
The extreme value of the equation is (-5,-4)
