1. Tangent Lines
Section 2.1

2. Secant Line
A secant line is a line that connects two points on a graph.
Notice the slopes of secant lines are different depending which two points you connect.

3. Secant Line Up to this point we have used the formula
to find the slope of the secant line joining points (x1, y1) and
(x2, y2).
This formula has four inputs x1 and x2, y1, y2.. Once we input these four values into the formula, our output represents the slope of that secant line.

4. Slope of Secant Line
We will now find another formula for the slope of the secant line between two points.
We will rename the points (x1, y1) and (x2, y2) using function notation in order to obtain our new slope formula.
(x2, y2)
(x1, y1)

5. We will let x1 = x.
Then it follows that y1, the y-value for x1, can be rewritten in function notation as f(x).
Note that f(x) refers to the output when the input is x.
Similarly, we rename x2 as x + h, where h refers to the distance from x1 to x2.
Then using function notation, y2 will be rewritten as f(x + h), the output when the input is x + h.
Point (x1, y1) becomes (x, f(x)) and Point (x2, y2) becomes (x + h, f(x + h))

6. The slope formula becomes x1 x2 Simplify the Denominator
Recall that h is the distance from x1 to x2, namely x2 – x1

7. Note: This formula for the slope has only two inputs: x, the smallest x-value and h, where h is the distance from the first x-value to the second x-value.
Example: Find the formula for the slope of any secant line for the function .
Step 1: Find
Step 2: Substitute into the formula:
This is a function whose output is slope of a secant line for
inputs x (the first x coordinate) and h (the distance between the x’s).

8. For example:
Find the slope of the secant line to the graph of from x = -1 to x = 2.
We know the slope of the secant lines of the function f(x) =-2x2 follow the formula:
In this case, x = -1 (first x value from left to right) and h = 3 (distance from x = -1 to x = 2).
Slope = -2
Therefore, the slope of the secant line shown is = -4(-1) – 2(3) = 4 – 6 = -2

9. Tangent Line
Tangent Line: The tangent line is a line drawn at a single point on a graph.
How do you draw a tangent line at an x-value? Think of having a rock at the end of a string and following some curve with this rock.
If you release the string at a point, say at x = ½, the path the rock follows is your tangent line at x = ½ .
Likewise, the path the rock follows if released at x = 3 would be the tangent line drawn at x = 3.
A tangent line can be drawn at each point.

10. Finding the slope of the tangent line.
Tangent line at x = -1.
How do we find the slope of the tangent line, say at x = -1 for the graph of f(x) = x3 – 3x2?
Notice we cannot use the
formula
because it would require two points on the line. We only know one point, the point of tangency (-1, -4)

11. because we do not have h, the distance b/w the two x-values.
We cannot use
because we do not have h, the distance b/w the two x-values.
Tangent line at x = -1.
Again let’s allow the second x-value get closer to x = -1 and draw the secant line.
Again let’s allow the second x-value get closer to x = -1 and draw the secant line.
Let’s draw a secant line from x = -1 to x = any other x- value.
Secant line.
Obviously, the slope of this secant line is different from the slope of our tangent line.
Notice that although the secant line is different from the tangent line, they are getting closer together as the second x-value gets closer to x = -1.
Let’s allow the second x-value get closer to x = -1 and draw the secant line again.

12. If we continue to allow the second x-value closer to x = -1 then the secant line will approach the tangent line.
Notice that the second point approaching x = -1 simply means that the distance b/w the two x-values is approaching 0. (h 0)

13. Therefore, to find the slope of the tangent line…
we find the slope of the
secant line,
then take the limit as h0

14. Example-Polynomial
Find the slope of the tangent line to f(x) = x3 – 3x2 at x = -1.

15. Given the equation
a) Find the slope of a secant line through the given point

16. Given the equation
b) Find the SLOPE of a tangent line through the given point
From part a,
At (4, 0), 2x – 4 = 2(4) – 4 = 4
c) Find the EQUATION of a tangent line through the given point
From part b, we have the slope (4)….and we have the pt (4, 0)
y – 0 = 4(x – 4)
c) Find the value of x for which the slope of the tangent line is 0
2x – 4 = 0
x = 2

17. Given the equation
a) Find the slope of a secant line through the given point

18. Given the equation
b) Find the SLOPE of a tangent line through the given point
From part a,
At (1, -1),
c) Find the EQUATION of a tangent line through the given point
From part b, we have the slope (1)….and we have the pt (1, -1)
y + 1 = 1(x - 1)
c) Find the value of x for which the slope of the tangent line is 0

19. Using the limit definition, find the first derivative of

20. Using the limit definition, find the first derivative of

21. Using the limit definition, find the first derivative of