2Secant LineA 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.
3Secant 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.
4Slope of Secant LineWe 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)
5We 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))
6The slope formula becomes x1 x2 Simplify the Denominator Recall that h is the distance from x1 to x2, namely x2 – x1
7Note: 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: FindStep 2: Substitute into the formula:This is a function whose output is slope of a secant line forinputs x (the first x coordinate) and h (the distance between the x’s).
8For 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 = -2Therefore, the slope of the secant line shown is = -4(-1) – 2(3) = 4 – 6 = -2
9Tangent LineTangent 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.
10Finding 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 theformulabecause it would require two points on the line. We only know one point, the point of tangency (-1, -4)
11because we do not have h, the distance b/w the two x-values. We cannot usebecause 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.
12If 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)
13Therefore, to find the slope of the tangent line… we find the slope of thesecant line,then take the limit as h0
14Example-PolynomialFind the slope of the tangent line to f(x) = x3 – 3x2 at x = -1.
15Example-Rational Function Find the equation of the tangent line to y = 2/x at x = 2.Point of Tangency
16Rate of ChangeThe slope of a secant line (line that connects two points on a graph) is sometimes referred to as an average rate of change.Over an intervalratioVertical change vs. horizontal change.One quantity that is an average rate of change is the average velocity ( vavg).Velocityave =
17Example-Position Function The position of a car after t minutes driving in a straight line is given byThis formula will give you the position of the car on a number line (output) given the time in minutes (input) fort0 (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)
18The position of a car after t minutes driving in a straight line is given by 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.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 8/3.5/125/65/45/325/125/235/1210/3
19Position vs Time curve Find the average velocity from x = 2 to x = 4. Position on number lineFind the average velocity from x = 2 to x = 4.Slope = Average velocity from x = 2 to x = 4.Looking for the change in position with respect to the time interval x = 2 to x = 4.Time in minutesLooking for the slope of the secant line from x = 2 to x = 4.
20Example-Average Velocity The position of a car after t minutes driving in a straight line is given byFind the average velocity from x = 2 to x = 4.
21Instantaneous rate of change = Change in y at an instant of time =slope of the tangent line at a point.
22Example-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.
23Example-Estimating the slope of a tangent line See page 156, # 6