What is the slope of the tangent line to the graph at f(x)=x^2-5 at the point (2,-1)?
f(x)=x^sup2;-5
Use the power rule to find f'(x)
=>
f'(x)=2x
at point P(2,-1),
x=2
therefore
f'(x)=f'(2)=2*2=4
Slope of the tangent equals the derivative of the function at the point in question.
To find the slope of the tangent line to the graph of f(x) = x^2 - 5 at the point (2, -1), we need to take the derivative of the function and evaluate it at x = 2.
Step 1: Find the derivative of f(x)
To find the derivative of f(x), we'll need to use the power rule. For any function of the form f(x) = x^n, the derivative is given by f'(x) = n*x^(n-1).
In this case, we have f(x) = x^2 - 5. Taking the derivative, we get:
f'(x) = 2*x^(2 - 1) = 2*x.
Step 2: Evaluate the derivative at x = 2
To find the slope of the tangent line, we need to evaluate the derivative at x = 2. Plugging in x = 2 into the derivative we found in Step 1, we get:
f'(2) = 2*2 = 4.
Therefore, the slope of the tangent line to the graph of f(x) = x^2 - 5 at the point (2, -1) is 4.