How do you find the horizontal tangent for the equation f(x)=x^2-3?

Question

Algebraically

Answer 1

take the derivative to get the slope
f'= 2x

so that is the slope of the tangent at any given x

Answer 2

f'(x) = 2x

at a horizontal tangent, f'(x) = 0
so 2x=0
x=0
f(0) = 0-3 = -3

so the horizontal tangent is y = -3

Answer 3

where did you get f(x)=2x??

Answer 4

it is hard to read but it says
f '(x) = 2x , which is short form for the derivative of f(x)

I assumed you know Calculus.
This is a Calculus-type question