How to find the vertex ?

Question

f(x)= x^2-3

Answer 1

x^2 is never less than zero, so f(x) is never less than -3. The vertex is at (0,-3)

In general, f(x) = ax^2+bx+c has its vertex at x = -b/2a

f(x) = a(x-h)^2 + k

has its vertex at (h,k). In this case, h=0 and k = -3, so the vertex is at (0,-3)

Answer 2

If you do not do calculus you complete the square

x^2 - 3 = y

x^2 - 3 + (3/2)^2 = y + 9/4

(x-3/2)^2 = y+9/4
vertex at (3/2 , -9/4)

Answer 3

assumed x with 3