Download presentation

Presentation is loading. Please wait.

Published byBarbara Joynt Modified about 1 year ago

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 (x 1, y 1 ) and (x 2, y 2 ). This formula has four inputs x 1 and x 2, y 1, y 2.. 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 (x 1, y 1 ) and (x 2, y 2 ) using function notation in order to obtain our new slope formula. (x 1, y 1 ) (x 2, y 2 )

5
We will let x 1 = x. Then it follows that y 1, the y-value for x 1, 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 x 2 as x + h, where h refers to the distance from x 1 to x 2. Then using function notation, y 2 will be rewritten as f(x + h), the output when the input is x + h. Point (x 1, y 1 ) becomes (x, f(x)) and Point (x 2, y 2 ) becomes (x + h, f(x + h))

6
The slope formula x1x1 Simplify the Denominator x2x2 Recall that h is the distance from x 1 to x 2, namely x 2 – x 1 becomes

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) =-2x 2 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). Therefore, the slope of the secant line shown is = -4(-1) – 2(3) = 4 – 6 = -2 Slope = -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. 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) 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) = x 3 – 3x 2 ?

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

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
Find the slope of the tangent line to f(x) = x 3 – 3x 2 at x = -1. Example-Polynomial

15
Example-Rational Function Point of Tangency Find the equation of the tangent line to y = 2/x at x = 2.

16
Rate of Change The slope of a secant line (line that connects two points on a graph) is sometimes referred to as an average rate of change. ratio Over an interval Vertical change vs. horizontal change. One quantity that is an average rate of change is the average velocity ( v avg ). Velocity ave =

17
Example-Position Function t 00 (positioned at 0 on the number line at the beginning) 15/12 (moved to 5/12 on the number line after 1 minute) 24/3 (moved to the right to position 4/3 on the number line after 2 minutes) 39/4 (moved to the right to position 9/4 on the number line after 3 min.) 48/3 (moved to position 8/3 on the number line after 4 minutes) The position of a car after t minutes driving in a straight line is given by This formula will give you the position of the car on a number line (output) given the time in minutes (input) for

18
0 The position of a car after t minutes driving in a straight line is given by 5/125/6 5/45/3 25/12 5/2 35/1210/3 At t = 0, the position is 0. At t = 1, the position is 5/12. At t = 2, the position is 4/3. At t = 3, the position is 9/4. At t = 4, the position is 8/3. We can also graph the time vs. position graph on an x-y plane. The x would represent the time (min) and y would represent the position on the number line.

19
Position vs Time curve Time in minutes Position on number line Looking for the change in position with respect to the time interval x = 2 to x = 4. Looking for the slope of the secant line from x = 2 to x = 4. Slope = Average velocity from x = 2 to x = 4. Find the average velocity from x = 2 to x = 4.

20
The position of a car after t minutes driving in a straight line is given by Example-Average Velocity Find the average velocity from x = 2 to x = 4.

21
Instantaneous rate of change = Change in y at an instant of time = slope of the tangent line at a point.

22
Example-Instantaneous Rate of Change (Instantaneous Velocity) Given the position of an object moving along a straight line is, find the instantaneous rate of change or the instantaneous velocity at t = 2.

23
Example-Estimating the slope of a tangent line See page 156, # 6

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google