A line that is tangent to a function f(:c) at 33 = a is a line that touches it") at a: : (13 whose slope is the same as the instantaneous rate of change of f(:r) at J: = a. 2. Do your best to sketch the line tangent to at) at the point (1,U) on your graph. 3. Approximate the slope of your tangent line. Recall the concept of a secant line. 4. Explain what it means for a line to be secant to f(:t) 5. You will sketch three secant lines on your graph. Each line should go through the point (1,0). A second point on the line will be: (i) (011%)) = (0) (a) (0.5, 1105)) = (0.5; ) (iii) sugar0.9)) = (0.9= J 6. Find the slope of each of these lines: 7' How do the slopes of the secant lines compare to the slope of the tangent line? Explain. 8. Find the slope of the secant line through the point (1,(]) and (1 + h, f(1 + 31)) 9. How could you use the slope you found in # 8 to nd the slope of your tangent line? 10. Execute your idea from 7% 9. 11. What is the slope of your tangent line? How did it compare to your guess from # 3? Suppose that we wanted to find the slopes of several different tangent lines. We could repeat our previous steps for each tangent line, or we could just do it once, in general. Suppose that we want to find the slope of the line tangent to f(x) at the point (a, f(a)). (Then, we can plug in any a value that we want!) 12. Find the slope of the secant line through the points (a, f(a)) and (ath, f(a + h)).13. Explain how this will help you to find the slope of the line tangent to f(x) at the point (a, f(a)). 14. Using your answers from # 12 and # 13, find the slope of the line tangent to f(x) at the point (a, f(a)). 15. Use your answer to # 14 to find the slope of the line tangent to f(x) at the given point. Draw a sketch in each case (graph paper attached). (i) The line tangent to f(x) at the point (1, f(1)) has slope (ii) The line tangent to f(x) at the point (0, f(0)) has slope (iii) The line tangent to f(x) at the point (-1, f(-1)) has slope