How do you find the horizontal tangent for the equation f(x)=x^2-3?
Algebraically
take the derivative to get the slope
f'= 2x
so that is the slope of the tangent at any given x
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
where did you get f(x)=2x??
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
To find the horizontal tangent for the equation f(x) = x^2 - 3 algebraically, you need to find the derivative of the function with respect to x. The derivative will tell us the slope of the tangent line at any given point on the curve.
Step 1: Start with the original function: f(x) = x^2 - 3.
Step 2: Differentiate f(x) with respect to x. For this specific function, we can use the power rule of differentiation, which states that the derivative of x^n is n*x^(n-1). Applying this rule, we get:
f'(x) = 2x.
So the derivative of f(x) is f'(x) = 2x.
Step 3: Set the derivative f'(x) equal to zero and solve for x. This is because the slope of a horizontal line is always zero. Thus, when we find the values of x that make the derivative equal to zero, we will have the x-coordinates of the points where the tangent line is horizontal.
2x = 0
Divide both sides of the equation by 2:
x = 0
Step 4: Since x = 0 is the only solution, the horizontal tangent line occurs when x = 0.
So, the equation of the horizontal tangent line is y = f(0). Substituting x = 0 into the original function f(x), we get:
y = 0^2 - 3 = -3.
Therefore, the equation of the horizontal tangent line for the function f(x) = x^2 - 3 is y = -3.
Note: The x-coordinate (0 in this case) represents the point at which the horizontal tangent line intersects the curve. You can also verify this visually by graphing the function and observing the point of intersection.