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- 1. Calculus AB Review Slopes of non-linear secant/tangent lines, and point-slope formula Kenyon HundleyTuesday, September 25, 12
- 2. Important Formulas Slope of a secant line at two points: f(x₀)-f(x₁) m=-------- x₀-x₁ Slope of a tangent line at one point: f(h+x₀)-f(x₀) m=-------- h Point-Slope Equation: y-y₀=m(x-x₀)Tuesday, September 25, 12
- 3. x-2 1. f(x)=--- x-5 a) find the slope of the secant line between x = 3 and x = 4 b) find the slope of the tangent line at x = 2 c) find the equation of the tangent line at x = 2Tuesday, September 25, 12
- 4. 1.a) find the slope of the secant line between x = 3 and x = 4 Our Line’s Equation x-2 f(x)=--- x-5 Secant Line Slope Formula f(x₀)-f(x₁) m=-------- x₀-x₁Tuesday, September 25, 12
- 5. 1.a) Steps 1. Using the secant line slope formula we can plug in the two values of ‘x’ that are given to us. We’ll just say x₀=3 and x₁=4. This looks like: f(3)-f(4) m=-------- 3-4 2. So on the top we see we have f(3) and f(4), which we interpret as “plug in 3 and 4 as x- values into the function and the solve for the y-value (or f(x) in other words)”. When we plug 3 and 4 in as x-values to our given equation, it looks like this: 3-2 4-2 f(3)=--- = -1/2 f(4)=--- = -2 3-5 4-5Tuesday, September 25, 12
- 6. 1.a) Steps 3. So now that we know f(3)=-1/2 and f(4)=-2, we can plug these values back into our secant slope equation which will look like: (-1/2)-(-2) m=-------- 3-4 4. So from here we’ll do basic simplification, which results in our answer: m = -1.5Tuesday, September 25, 12
- 7. 1.b) find the slope of the tangent line at x = 2 Our Line’s Equation x-2 f(x)=--- x-5 Tangent Line Slope Formula f(h+x₀)-f(x₀) m=-------- hTuesday, September 25, 12
- 8. 1.b) Steps 1. Using the tangent line slope formula we’ll plug in the value of ‘x’ that is given to us. Since x=2, this looks like: f(2+h)-f(2) m=-------- h 2. So in order to find the slope, we want to simplify the equation to a point where, if we plugged 0 in for ‘h’, we would not result in an undefined equation (0 as the denominator). We’ll start by solving the functions that are defined in the numerator, f(2+h) and f(2): 2+h-2 2-2 f(2+h)= --- = h/(h-3) f(2)=--- = 0 2+h-5 2-5Tuesday, September 25, 12
- 9. 1.b) Steps 3. So, now we’ll look at the equation with the function values plugged in (f(h+2)=h/(h-3), f(2)=0). h/(h-3) h 1 m=-------- = h -------- = -------- h(h-3) h-3 4. Now with our equation m=1/(h-3), we can see that plugging in 0 for ‘h’ won’t result in an undefined value so we can use this to solve for the slope. h 1 m = -- = -1/3 as h approaches 0 0-3 We write the answer like “m = -1/3 as h approaches 0” because we cannot find the exact slope of the tangent line at a given point, but we can get really close to it by plugging in 0 for ‘h’, or a decimal value extremely close to 0.Tuesday, September 25, 12
- 10. 1.c) find the equation of the tangent line at x = 2 Our Line’s Equation x-2 f(x)=--- x-5 Point-Slope Formula y-y₁ = m(x-x₁) Tangent Line’s Slope -1/3Tuesday, September 25, 12
- 11. 1.c) Steps 1. To begin solving for the tangent line’s equation, we substitute x=2 in to our line’s equation: 2-2 f(x)= --- = 0 2-5 2. So now we know that 0 is x’s corresponding y- value. We’ll use (2,0) as our x₁ & y₁ in the point-slope equation, and also -1/3 as the slope (we found the slope of the same tangent line in 1.b): y-0 = -1/3(x-2)Tuesday, September 25, 12
- 12. 1.c) Steps 3. Now we basically just simplify the equation to slope-intercept form, then we have our answer: y = -1/3x+2/3Tuesday, September 25, 12
- 13. 2. f(x)=2x²-3x+1 a) find the slope of the secant line between x = -1 and x = 2 b) find the slope of the tangent line at x = 2 c) find the equation of the tangent line at x = 2Tuesday, September 25, 12
- 14. 2.a) Steps 1. Using the secant line slope formula we can plug in the two values of ‘x’ that are given to us. We’ll just say x₀=-1 and x₁=2. This looks like: f(-1)-f(2) m=-------- 3-4 2. So on the top we see we have the functions f(-1) and f(2). When we plug -1 and 2 in as x-values to our given equation, it looks like this: f(-1)=2(-1)²-3(-1)+1= 6 f(2)=2(2)²-3(2)+1= 3Tuesday, September 25, 12
- 15. 2.a) Steps 3. So now that we know f(-1)=6 and f(2)=3, we can plug these values back into our secant slope equation which will look like: (6)-(3) m=-------- -1-2 4. So from here we’ll do basic simplification, which results in our answer: m = -1Tuesday, September 25, 12
- 16. 2.b) find the slope of the tangent line at x = 2 Our Line’s Equation f(x)=2x²-3x+1 Tangent Line Slope Formula f(h+x₀)-f(x₀) m=-------- hTuesday, September 25, 12
- 17. 2.b) Steps 1. Using the tangent line slope formula we’ll plug in the value of ‘x’ that is given to us. Since x=2, this looks like: f(2+h)-f(2) m=-------- h 2. So in order to find the slope, we want to simplify the equation to a point where, if we plugged 0 in for ‘h’, we would not result in an undefined equation (0 as the denominator). We’ll start by solving the functions that are defined in the numerator, f(2+h) and f(2): f(2+h)=2(2+h)²-3(2+h)+1=2h²+3+5h f(2)=2(2)²-3(2)+1=3Tuesday, September 25, 12
- 18. 2.b) Steps 3. So, now we’ll look at the equation with the function values plugged in. 2h²+3+5h-3 2h²+5h 2h+5 m=-------- = h -------- = -------- h 1 4. Now with our equation m=2h+5, we can see that plugging in 0 for ‘h’ won’t result in an undefined value so we can use this to solve for the slope. h m = 0+5 = 5 as h approaches 0Tuesday, September 25, 12
- 19. 2.c) find the equation of the tangent line at x = 2 Our Line’s Equation f(x)=2x²-3x+1 Point-Slope Formula y-y₁ = m(x-x₁) Tangent Line’s Slope 5Tuesday, September 25, 12
- 20. 2.c) Steps 1. To begin solving for the tangent line’s equation, we substitute x=2 in to our line’s equation: y=2(2)²-3(2)+1 = 3 2. So now we know that 3 is x’s corresponding y- value. We’ll use (2,3) as our x₁ & y₁ in the point-slope equation, and also 5 as the slope (we found the slope of the same tangent line in 2.b): y-3 = 5(x-2)Tuesday, September 25, 12
- 21. 2.c) Steps 3. Now we basically just simplify the equation to slope-intercept form, then we have our answer: y = 5x-7Tuesday, September 25, 12
- 22. 3. If an object travels a distance of s = 2t²-5t+1, where s is in feet and t is in seconds,find: a) the average velocity of object within the first 10 seconds. b) the instantaneous velocity of the object at t = 3 seconds.Tuesday, September 25, 12
- 23. 3.a) the average velocity of object within the first 10 feet Our Line’s Equation s(t)= 2t²-5t+1 Velocity Average (Secant Line Slope) Formula s(t₀)-s(t₁) Vavg=-------- t₀-t₁Tuesday, September 25, 12
- 24. 3.a) Steps 1. In this problem, we basically use the same rules and formula as the secant line slope formula, except we use different notation. We can plug 10(seconds) and 0(seconds) for t₀ and t₁ in order to find our average velocity during the first ten seconds. We’ll say t₀=0 and t₁ =10. This looks like: s(0)-s(10) Vavg=-------- 0-10 2. So on the top we, once again, see that we have functions; s(0) and s(10) are our functions. After plugging them into our line’s equation, we get this: s(0)= 2(0)²-5(0)+1 = 1 s(10)= 2(10)²-5(10)+1= 151Tuesday, September 25, 12
- 25. 3.a) Steps 3. So now that we know s(0)=1 and f(10)=151, we can plug these values back into our average velocity equation which will look like: (1)-(151) Vavg=-------- 0-10 4. So from here we’ll do basic simplification, which results in our answer: Vavg = 15 feet per secondTuesday, September 25, 12
- 26. 3.b) the instantaneous velocity of the object at t = 3 seconds Our Line’s Equation s(t)= 2t²-5t+1 Instantaneous Velocity (Tangent Line Slope) Formula s(h+t₀)-s(t₀) Vinst= --------- hTuesday, September 25, 12
- 27. 3.b) Steps 1. To discover the instantaneous velocity at t=3 seconds, we basically use the tangent line slope formula but with different notation. We plug in the given value of 3 in for “t” in the Vinst equation. This looks like: s(3+h)-s(3) Vinst=-------- h 2. So, (like the tangent line slope equations) in order to find the instantaneous velocity, we want to simplify the equation to a point where, if we plugged 0 in for ‘h’, we would not result in an undefined equation (0 as the denominator). To start we’ll solve the functions that are defined in the numerator, s(3+h) and s(3): s(h+3)= 2(h+3)²-5(h+3)+1= 2h²+7h+4 s(3)= 2(3)²-5(3)+1= 4Tuesday, September 25, 12
- 28. 3.b) Steps 3. So, now we’ll look at the equation with the function values plugged in. 2h²+7h+4-4 2h²+7h 2h+7 Vinst=-------- = h -------- = -------- h 1 4. Now with our equation Vinst=2h+7, we can see that plugging in 0 for ‘h’ won’t result in an undefined value so we can use this to solve for the instantaneous velocity at 3 seconds. Vinst = 0+7 = 7 feet per secondTuesday, September 25, 12

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