Find the slope of the tangent line to the graph y = x^-3 at the point (0.5, 8).
-48
-12
4
16
y'=-3/x^2 = -3(4)
The derivative should have been
dy/dx = -3x^-4
= -3/x^4
when x = .5 or 1/2
dy/dx = -3/(1/2)^4
= -3/(1/16
= -3(16)
= -48
To find the slope of the tangent line to the graph at a specific point, you need to find the derivative of the function at that point. In this case, we are given the function y = x^-3 and the point (0.5, 8).
Step 1: Find the derivative of the function y = x^-3.
To find the derivative of y = x^-3, we need to use the power rule. The power rule states that if we have a function of the form y = x^n, then its derivative is given by dy/dx = nx^(n-1).
So, the derivative of y = x^-3 is dy/dx = -3x^(-3-1) = -3x^-4 = -3/x^4.
Step 2: Evaluate the derivative at the given point.
To find the slope of the tangent line at the point (0.5, 8), we substitute x = 0.5 into the derivative function we found in Step 1.
dy/dx = -3/(0.5)^4 = -3/0.0625 = -48.
Step 3: Interpret the result.
The slope of the tangent line to the graph y = x^-3 at the point (0.5, 8) is -48.
Therefore, the correct answer is -48.