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Tangents and Slopes of Curves

Find equations of tangents and slopes of curves at given points using calculus methods. Explore different curves and their properties.

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Tangents and Slopes of Curves

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  1. 1. The curve y = 3x2 − 5x − 11 is shown in the diagram. Find (i)

  2. 1. The curve y = 3x2 − 5x − 11 is shown in the diagram. Hence, find the slope of the tangent to the curve at the point where x = 2. (ii) Therefore, the slope of the tangent = 7

  3. 2. (i) If y = x3 + x2 − 12x, find Hence, find the slope of the tangent to the curve at the point where:

  4. 2. (ii) x = 1 Therefore, the slope of the tangent = −7

  5. 2. (iii) x = −3 Therefore, the slope of the tangent = 9

  6. 3. The curve y = −x2 + 5x − 7 is shown in the diagram. Find (i)

  7. 3. The curve y = −x2 + 5x − 7 is shown in the diagram. Find the slope of the tangent to the curve at the point (4, − 3). (ii)

  8. 3. The curve y = −x2 + 5x − 7 is shown in the diagram. Hence, find the equation of this tangent. (iii)

  9. 4. (i) If f (x) = 3x2 + 5x + 4, find f ʹ (x) f (x) = 3x2 + 5x + 4 f ' (x) = 6x + 5

  10. 4. (ii) Hence, find the equation of the tangent to the curve at the point (−1, 2). Slope = f ' (−1) = 6(−1) + 5 Slope = −6 + 5 Slope = −1 (−1, 2) m = −1 x1, y1 Equation: y − y1 = m(x − x1) y − 2 = −1(x − (− 1)) y − 2 = −1(x + 1) y − 2 = − x − 1 x + y − 1 = 0

  11. 4. (iii) Find the coordinates of the points where this tangent intercepts the x and y axes. x + y − 1 = 0 At the y-axis, x = 0: At the x-axis, y = 0: x + y − 1 = 0 x + y − 1 = 0 0 + y − 1 = 0 x + 0 − 1 = 0 y − 1 = 0 x − 1 = 0 y = 1 x = 1 (0, 1) (1, 0)

  12. 5. Find the equation of the tangent to the curve f (x) = 5x2 − 3x − 7, at the point (2, 7). (2, 7) m = 17 f (x) = 5x2 − 3x − 7 x1, y1 Slope = f ′(x) = 10x − 3 Equation: y − y1 = m(x − x1) x = 2: f ′(2) = 10(2) − 3 y − 7 = 17(x − 2) Slope = 20 − 3 y − 7 = 17x − 34 Slope = 17 17x − y − 34 + 7 = 0 17x − y − 27 = 0

  13. 6. Find the equation of the tangent to the curve g(x) = 2x3 + x2 − 23x + 20, at the point (2, − 6). g(x) = 2x3 + x2 − 23x + 20 (2, − 6) m = 5 x1, y1 g′(x) = 6x2 + 2x − 23 Equation: y − y1 = m(x − x1) Slope at x = 2: y −(− 6) = 5(x − 2) g′(2) = 6(2)2 + 2(2) − 23 y + 6 = 5x − 10 = 24 + 4 − 23 5x − y − 10 − 6 = 0 = 5 5x − y − 16 = 0

  14. 7. Find the equation of the tangent to the curve f (x) = −2x2 + 7x − 5, at the point where x = 1. f(x) = −2x2 + 7x − 5 f ′ (x) = −4x + 7 Slope = f ′ (1) = −4(1) + 7 = − 4 + 7 = 3

  15. 7. Find the equation of the tangent to the curve f (x) = −2x2 + 7x − 5, at the point where x = 1. To find y-value: (1, 0) m = 3 x1, y1 y = −2x2 + 7x − 5 x = 1: y = −2(1)2 + 7(1) − 5 Equation: y − y1 = m(x − x1) y = −2 + 7 − 5 y − 0 = 3(x − 1) y = 0 y = 3x − 3 Point (1, 0) 3x − y − 3 = 0

  16. 8. The graph of f (x) = 5x2 − 2x + 1 is shown in the diagram. Find f ′(x). (i) f (x) = 5x2 − 2x + 1 f ′ (x) = 10x – 2

  17. 8. The graph of f (x) = 5x2 − 2x + 1 is shown in the diagram. Find the equation of the tangent, n, to the curve at the point where x = 1. (ii) Slope: f ′ (x) = 10x − 2 When x = 1: f ′ (x) = 10(1) − 2 f ′ (x) = 10 − 2 Slope = 8

  18. 8. The graph of f (x) = 5x2 − 2x + 1 is shown in the diagram. Find the equation of the tangent, n, to the curve at the point where x = 1. (ii) Find y coordinate: y = 5x2 − 2x + 1 y = 5(1)2 − 2(1) + 1 y = 5 − 2 + 1 y = 4 Point: (1, 4)

  19. 8. The graph of f (x) = 5x2 − 2x + 1 is shown in the diagram. Find the equation of the tangent, n, to the curve at the point where x = 1. (ii) (1, 4) m = 8 x1, y1 Equation: y − y1 = m(x − x1) y − 4 = 8(x − 1) y − 4 = 8x − 8 8x − y − 8 + 4 = 0 n : 8x − y − 4 = 0

  20. 8. The graph of f (x) = 5x2 − 2x + 1 is shown in the diagram. Find the equation of the tangent, k, to the curve at the point where x = −2. (iii) Slope: f ′ (x) = 10x − 2 When x = −2: f ' (−2) = 10(−2) – 2 f ' (−2) = − 20 − 2 Slope = −22

  21. 8. The graph of f (x) = 5x2 − 2x + 1 is shown in the diagram. Find the equation of the tangent, k, to the curve at the point where x = −2. (iii) Find y coordinate: y = 5x2 − 2x + 1 y = 5(−2)2 − 2(−2) + 1 y = 20 + 4 + 1 y = 25 Point: (−2, 25)

  22. 8. The graph of f (x) = 5x2 − 2x + 1 is shown in the diagram. Find the equation of the tangent, k, to the curve at the point where x = −2. (iii) (−2, 25) m = −22 x1, y1 Equation: y − y1 = m(x − x1) y − 25 = −22(x + 2) y − 25 = −22x − 44 22x + y − 25 + 44 = 0 22x + y + 19 = 0

  23. 9. The diagram shows two tangents to the curve h(x) = x2 − 4x + 2 at the points (1, −1) and (3, −1). Find the equations of these tangents. (i) (a) (1, −1) h(x) = x2 − 4x + 2 Slope = h′ (x) = 2x − 4 Slope when x = 1: h′ (1) = 2(1) – 4 = 2 − 4 Slope = −2

  24. 9. The diagram shows two tangents to the curve h(x) = x2 − 4x + 2 at the points (1, −1) and (3, −1). Find the equations of these tangents. (i) (a) (1, −1) (1, −1) m = −2 x1, y1 Equation: y − y1 = m(x − x1) y − (−1) = −2(x − 1) y + 1 = −2x + 2 2x + y + 1 − 2 = 0 2x + y − 1 = 0

  25. 9. The diagram shows two tangents to the curve h(x) = x2 − 4x + 2 at the points (1, −1) and (3, −1). Find the equations of these tangents. (i) (b) (3, −1) Slope when x = 3: hʹ(x) = 2x − 4 hʹ(3) = 2(3) − 4 Slope = 6 − 4 Slope = 2

  26. 9. The diagram shows two tangents to the curve h(x) = x2 − 4x + 2 at the points (1, −1) and (3, −1). Find the equations of these tangents. (i) (b) (3, −1) (3, −1) m = 2 x1, y1 Equation : y − y1 = m(x − x1) y − (−1) = 2(x − 3) y + 1 = 2x – 6 2x − y − 6 − 1 = 0 2x − y − 7 = 0

  27. 9. The diagram shows two tangents to the curve h(x) = x2 − 4x + 2 at the points (1, −1) and (3, −1). Investigate if the two tangents are perpendicular to one another. (ii) If the tangents are perpendicular, the product of their slopes equals −1 Slope of tangent at (1, −1) = −2 Slope of tangent at (3, −1) = 2 Product= (−2)(2) = −4 ≠ −1 Therefore, tangents are not perpendicular.

  28. 9. The diagram shows two tangents to the curve h(x) = x2 − 4x + 2 at the points (1, −1) and (3, −1). Find the point of intersection of the two tangents. (iii) 2x + y − 1 = 0 (a) 2x − y − 7 = 0 (b) Solve these equations simultaneously: 2x + y − 1 = 0 2x − y − 7 = 0 (add the rows to eliminate the y’s) 4x − 8 = 0 4x = 8 x = 2

  29. 9. The diagram shows two tangents to the curve h(x) = x2 − 4x + 2 at the points (1, −1) and (3, −1). Find the point of intersection of the two tangents. (iii) 2x + y − 1 = 0 (a) 2x − y − 7 = 0 (b) Substitute x = 2 into: 2x + y − 1 = 0 2(2) + y − 1 = 0 4 + y − 1 = 0 y + 3 = 0 y = −3 Point of intersection (2, −3)

  30. 10. (i) If f (x) = −3x2 + 5x + 12, find f ′(x). Hence, f (x) = −3x2 + 5x + 12 f ′(x) = −6x + 5

  31. 10. (ii) Find the equation of the tangent, t, to the curve, at the point (1, 14) (1, 14) m = −1 Slope = f ′(x) = −6x + 5 x1, y1 Slope when x = 1: Equation: y − y1 = m(x − x1) f ′(1) = −6(1) + 5 y − 14 = −1(x − 1) = −6 + 5 y − 14 = − x + 1 Slope = −1 x + y − 14 − 1 = 0 x + y − 15 = 0

  32. 10. (iii) find the equation of the tangent, s, to the curve, at the point (−1, 4) (−1, 4) m = 11 Slope when x = −1: x1, y1 f ′(−1) = −6(−1) + 5 Equation: y − y1 = m(x − x1) = 6 + 5 y − 4 = 11(x + 1) Slope = 11 y − 4 = 11x + 11 11x − y + 11 + 4 = 0 11x − y + 15 = 0

  33. 10. (iv) find the point of intersection of the tangentss and t. x + y − 15 = 0 (a) 11x − y + 15 = 0 (b) Solve these equations simultaneously: x + y − 15 = 0 11x − y + 15 = 0 (add the rows to eliminate the y’s) 12x = 0 x = 0 substitute x = 0 into x + y − 15 = 0 0 + y − 15 = 0 y = 15 Point of intersection = (0, 15)

  34. 10. (v) sketch the function f (x) and its tangents, tand s. At the x-axis, y = 0 −3x2 + 5x + 12 = 0 (multiply both sides by − 1) 3x2 − 5x − 12 = 0 (3x + 4)(x − 3) = 0 3x + 4 = 0 x − 3 = 0 3x = −4 x = 3 x =

  35. 11. (i) If f ′(x) = x2 − 5x + 8, find f ′(x). f (x) = x2 − 5x + 8 f ′(x) = 2x – 5

  36. 11. (ii) Find the coordinates of the point, for which the slope of the tangent is zero. Slope is zero when f ′(x) = 0 2x − 5 = 0 2x = 5 x = 2⋅5 y = x2 − 5x + 8 = (2⋅5)2 − 5(2⋅5) + 8 = 6⋅25 − 12⋅5 + 8 = 1⋅75 Coordinates: (2⋅5, 1⋅75)

  37. 12. (i) If g(x) = 3x2 + 12x − 7, find g ′(x). g (x) = 3x2 + 12x − 7 g ′(x) = 6x + 12

  38. 12. (ii) Find the coordinates of the point, for which the tangent is parallel to thex-axis (i.e. its slope is zero). Slope is zero g′(x) = 0 y = 3x2 + 12x − 7 6x + 12 = 0 = 3(−2)2 + 12(−2) − 7 6x = −12 = 12 − 24 − 7 x = −2 = −12 − 7 = −19 Co-ordinates (−2, −19)

  39. 13. Find the point on the curve y = 6x2 − 8x + 1, where its tangent is parallel to the line 4x – y – 7 = 0. Slope of 4x − y − 7 = 0 Rearrange the equation into the form y = mx + c − y = − 4x + 7 y = 4x – 7 Therefore, slope of this line is 4.

  40. 13. Find the point on the curve y = 6x2 − 8x + 1, where its tangent is parallel to the line 4x – y – 7 = 0. Slope of tangent to curve 6x2 − 8x + 1 Slope == 12x − 8 Let slope = = 4 12x − 8 = 4 12x = 4 + 8 12x = 12 (divide both sides by 12) x = 1

  41. 13. Find the point on the curve y = 6x2 − 8x + 1, where its tangent is parallel to the line 4x – y – 7 = 0. Find y coordinate: y = 6x2 − 8x + 1 Let x = 1 : y = 6(1)2 − 8(1) + 1 = 6 − 8 + 1 = −2 + 1 = −1 Point is (1, −1)

  42. 14. The diagram shows a sketch of the curve y = x3 − 3x − 1. The tangents at the points A and B are parallel to the x-axis. Find the coordinates of the points A and B. y = x3 − 3x − 1 Slope = = 3x2 – 3 at A and B, slope = 0: 3x2 − 3 = 0 (divide both sides by 3) x2 − 1 = 0 (x − 1)(x + 1) = 0 x = 1x = − 1

  43. 14. The diagram shows a sketch of the curve y = x3 − 3x − 1. The tangents at the points A and B are parallel to the x-axis. Find the coordinates of the points A and B. x = 1: y = x3 − 3x − 1 x = −1: y = x3 − 3x − 1 = (1)3 − 3(1) − 1 = (−1)3 − 3(−1) −1 = 1 − 3 − 1 = −1 + 3 −1 = −3 = 1 (1, −3) (−1, 1)

  44. 14. The diagram shows a sketch of the curve y = x3 − 3x − 1. The tangents at the points A and B are parallel to the x-axis. Find the coordinates of the points A and B. = 3x2 – 3 x = 1: 6 is positive, therefore, (1, −3) is the minimum point, B

  45. 14. The diagram shows a sketch of the curve y = x3 − 3x − 1. The tangents at the points A and B are parallel to the x-axis. Find the coordinates of the points A and B. x = −1: − 6 is negative, therefore, (− 1, 1) is the maximum point, A A(−1, 1) B(1, −3)

  46. 15. h(x) = x3 − 21x + 20 is a function. Find an expression for the slope of the tangent to the curve at any point x. (i) Slope of tangent to curve at any point x = h’(x) h(x) = x3 − 21x + 20 h' (x) = 3x2 − 21

  47. 15. h(x) = x3 − 21x + 20 is a function. Find the two values of x for which the slope of the tangent to the curve is 27. (ii) Slope = 27 3x2 − 21 = 27 3x2 = 27 + 21 3x2 = 48 (divide both sides by 3) x2 = 16 x = ± 4

  48. 15. h(x) = x3 − 21x + 20 is a function. Hence, find the coordinates of the two points where the slope of the tangent to the curve is 27. (iii) Find y coordinates: y = x3 − 21x + 20 y = x3 − 21x + 20 x = 4: y = (4)3 − 21(4) + 20 x = −4: y = (−4)3 − 21(−4) + 20 y = 64 − 84 + 20 y = −64 + 84 + 20 y = 0 y = 40 (4, 0) (−4, 40)

  49. 16. Find two points on the curve h(x) = x3 − 12x2 + 39x − 28, where the slope of the tangent is 3. h(x) = x3 − 12x2 + 39x − 28 h’(x) = 3x2 − 24x + 39 h′(x) = 3: 3x2 − 24x + 39 = 3 3x2 − 24x + 39 − 3 = 0 3x2 − 24x + 36 = 0 (divide both sides by 3) x2 − 8x + 12 = 0 (x − 6)(x − 2) = 0 x − 6 = 0 x − 2 = 0 x = 6 x = 2

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