Find the equation of the tangent line to the graph of f(x)= (8x^2-3)/(x^2+4) at the point where x=-1.

Take the derivative of f(x). That gives you m, the slope.

Then,

y= mx + b

Put in x,f(x) for x,y. Calculate b.

I get it now. Thanks for your help!

You're welcome! I'm glad I could help. Just to clarify, let me explain the steps in more detail on how to find the equation of the tangent line to the graph of f(x).

Step 1: Find the derivative of f(x)
To find the slope of the tangent line, we need to take the derivative of the function f(x). In this case, f(x) = (8x^2-3)/(x^2+4). To differentiate this function, we can use the quotient rule:

f'(x) = [(2(x^2+4)(16x) - (8x^2-3)(2x))/(x^2+4)^2]

Simplifying this expression will give you the derivative of f(x), which represents the slope of the tangent line at any given point.

Step 2: Substitute x = -1 into f'(x)
Next, we need to find the slope of the tangent line at x = -1. So, substitute x = -1 into the derivative f'(x) to evaluate the slope of the tangent line at that point.

Step 3: Find the equation of the tangent line
Now that we have the slope of the tangent line, we can use the point-slope form of a line equation, y = mx + b, where m is the slope and (x, y) represents a point on the line.

Since we know the point of tangency is x = -1, we can substitute -1 for x and the corresponding y-value you obtained from f(-1) into the equation.

Finally, you can solve for b, which is the y-intercept of the tangent line. Once you have m and b, you can write the equation of the tangent line in the form y = mx + b.

I hope this explanation helps! Let me know if you have any more questions.