Just as we can find the exact slope of the line tangent to this curve at...

 

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Just as we can find the exact slope of the line tangent to this curve at (1, -2) and (-3.-6) we should be able to find the slope of the tangent line at any other point on the curve. Instead of trying to do so at all of the points one-at-a-time, it's much easier to simply create a formula for all of them at once. We can do that by using a generic point (a, f (a)) on the curve. A point close to that point on the curve would be (a + h, f(a+h)) for a small value of h. Go Ay through the process you used in Problems 3 & 4 to find a formula for the slope a using the Ax points (a, f (a)) and (a + h. f(a+h)) at a. (This will take a fair amount of algebra! Show all algebra on a separate sheet of paper.) Then find the exact value of that slope at h = 0. This is f'(a) for any value a, the derivative of f at a which is the slope of the line tangent to the curve y = f(x) at (a, f (a)). f' (a) = Now fill in the table below using that formula: a 1 -3 5 -2 4 -6 f' (a) Just as we can find the exact slope of the line tangent to this curve at (1, -2) and (-3.-6) we should be able to find the slope of the tangent line at any other point on the curve. Instead of trying to do so at all of the points one-at-a-time, it's much easier to simply create a formula for all of them at once. We can do that by using a generic point (a, f (a)) on the curve. A point close to that point on the curve would be (a + h, f(a+h)) for a small value of h. Go Ay through the process you used in Problems 3 & 4 to find a formula for the slope a using the Ax points (a, f (a)) and (a + h. f(a+h)) at a. (This will take a fair amount of algebra! Show all algebra on a separate sheet of paper.) Then find the exact value of that slope at h = 0. This is f'(a) for any value a, the derivative of f at a which is the slope of the line tangent to the curve y = f(x) at (a, f (a)). f' (a) = Now fill in the table below using that formula: a 1 -3 5 -2 4 -6 f' (a)