How to find the vertex ?

f(x)= x^2-3

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)

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)

assumed x with 3

To find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula x = -b / (2a) to find the x-coordinate of the vertex. Once you have the x-coordinate, you can substitute it back into the function to find the corresponding y-coordinate.

In the given example, f(x) = x^2 - 3, we can see that a = 1, b = 0, and c = -3. Since b is 0 in this case, the formula becomes x = 0 / (2 * 1), which simplifies to x = 0.

Now we can substitute x = 0 back into the function to find the y-coordinate. Plugging in x = 0 into f(x) = x^2 - 3, we get f(0) = 0^2 - 3, which simplifies to f(0) = -3.

Therefore, the vertex of the function f(x) = x^2 - 3 is (0, -3).