Answer:
Equation in square form:
[tex]y=3(x+5)^2-4[/tex]
Extreme value:
[tex](h,k)=(-5,-4)[/tex]
Step-by-step explanation:
We are given
[tex]y=3x^2+30x+71[/tex]
we can complete square
[tex]y=3(x^2+10x)+71[/tex]
we can use formula
[tex]a^2+2ab+b^2=(a+b)^2[/tex]
[tex]y=3(x^2+2\times x\times 5)+71[/tex]
now, we can add and subtract 5^2
[tex]y=3(x^2+2\times x\times 5+5^2-5^2)+71[/tex]
[tex]y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71[/tex]
[tex]y=3(x+5)^2-75+71[/tex]
So, we get equation as
[tex]y=3(x+5)^2-4[/tex]
Extreme values:
we know that this parabola
and vertex of parabola always at extreme values
so, we can compare it with
[tex]y=a(x-h)^2+k[/tex]
where
vertex=(h,k)
now, we can compare and find h and k
[tex]y=3(x+5)^2-4[/tex]
we get
h=-5
k=-4
so, extreme value of this equation is
[tex](h,k)=(-5,-4)[/tex]
