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UNIT 1B LESSON 4

UNIT 1B LESSON 4. Slopes of Tangents. Slopes as Rate of Change. To begin our study of rates and changes we must realize that the rate of change is the change in y divided by the change in x . With a straight line, this has been the slope of the secant , m .

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UNIT 1B LESSON 4

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  1. UNIT 1B LESSON 4 Slopes of Tangents

  2. Slopes as Rate of Change To begin our study of rates and changes we must realize that the rate of changeis thechange inydivided by thechange inx. With a straight line, this has been theslope of the secant, m. We are assuming two points where x has been increased by an amount h Q P

  3. Approximating the slope of the tangent line with the secant line x means the change in x We sometimes use h instead of x P x + x x

  4. What happens as h or x gets smaller andsMALLER?

  5. or ∆x→0 Tangent to a Curve

  6. Examples of Tangents of Curves Since the curves are increasing and decreasing at different rates, we are looking for the instantaneousrate of changeat a particular point. This means we need the slope of thetangent lineat the point P.

  7. Tangent to a Curve • The process becomes: • 1. Start with what can be calculated, namely, the slope of a secant through P and a point Q nearby on the curve. • 2. Find the limiting value of the secant slope (if it exists) as Q approaches P along the curve. • 3. Define the slope of the curve at P to be this number and define the tangent to the curve at P to be the line through P with this slope.

  8. EXAMPLE on Page 1 of Unit 1 Lesson #4The point P(1, – 5) lies on the curve y = x2 – 6x. P(1, - 5) Find slope (m) at each point m = –5 –– 2.75 = – 4.5 1 – 0.5 m = –5 –– 4.59 = – 4.1 1 – 0.9 m = –5 –– 4.96 = – 4.0 1 – 0.99 - 4.0 m = –5 –– 5.04 = – 4.0 1 – 1.01 m = –5 –– 5.39 = – 3.9 1 – 1.1 Calculate values ofy with values of x very close to 1. m = –5 –– 6.75 = – 3.5 1 – 1.5

  9. Find the slope of the tangent line to the curve y = x2 – 6x, at P(1, – 5) using limits. m = lim f (x + h) – f (x) h→0 (x + h) – x m = lim [(x + h)2 – 6(x + h)] – [x2 – 6x] h→0 (x + h) – x m = lim x2 + 2xh + h2 – 6x – 6h – x2 + 6x h→0 h m = lim h(2x + h – 6) h→0 h Slope at x = 1 is 2(1) – 6 = –4 m = 2x – 6

  10. P(1, - 5) Find the equation of the tangent line to the curve y= x2 – 6x. at P(1, – 5) From previous work Check this result on your calculator Enter Y= x2 – 6x Graph. DRAW 2nd PRGM 5:Tangent Type in the x-value where you want the tangent line drawn.

  11. Complete Unit 1 Lesson #4 pages 2 through 6

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