Question

what is the minimum of f(x)=3x^2-12x-1? and how do i find it?

Answer 1

To find the value of the minimum of the function you can find the value of x that gives that value.

The value of x that give the minimum ( or the maximum, if a < 0) on a ax² + bx +c = 0 function it's:
\[x _{v}=\frac{ -b }{ 2a }\]
So, in that function, you have b = -12 and a = 3.
\[x _{v}=\frac{ -(-12) }{ 2*3 } = \frac{ 12 }{ 6 } = 2\]

Now you have to calc f(2).
\[f(2)=3*(2)^2-12*(2)-1=3*4-24-1=12-24-1=-13\]

Answer 2

thank you so much....