Needed help with this math question.

Question

f(x) = x^2 + 8x - 4

What is the minimum y - value y=?

Answer 1

Find the critical points by taking the derivative of this equation, y'=2x+8.  Set this equal to 0, since the minimum is the point where the slope at that point is 0.  (Remember that the physical meaning of the derivative at a point is the slope of the equation at that point.)  Solve for all possible values of x where y'=2x+8=0.  Subtracting 8 from both sides and then dividing 2 on each side, we get x=-4.

Plug this back into the initial equation to get f(x)=(-4)²+8(-4)-4.  This yields (-4, -20).