1 / 14

Curve Sketching

Curve Sketching. C1 Section 4.6 – 4.7. y = 9 - x 2. y = 2(9 - x 2 ). f( x ). 2f( x ). (2 , 5). (2 , 10). Each point twice as far along y. Stretches in the y-axis. If we have the function y = f( x ) then: y = a f( x ) is a stretch by a factor a in the y-axis:. f( x ). f(2 x ).

emmly
Télécharger la présentation

Curve Sketching

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Curve Sketching C1 Section 4.6 – 4.7

  2. y = 9 - x2 y = 2(9 - x2) f(x) 2f(x) (2 , 5) (2 , 10) Each point twice as far along y. Stretches in the y-axis • If we have the function y = f(x) then: • y = af(x) is a stretch by a factor a in the y-axis:

  3. f(x) f(2x) (3 , 0) (1.5 , 0) (2 , 5) (1 , 5) Stretches in the x-axis • If we have the function y = f(x) then: • y = f(ax) is a stretch by a factor1/ain the x-axis: y = 9 - x2 y = 9 – (2x)2 Each point 1/2 as far along x.

  4. (1 , 4) f(x) 0 (3, 1) Transformations of Points • f(x + a) Imagine a function where y = f(x), which has a root at 0, and points (1 , 4) and (3 , 1) lie on the curve: (0 , 4) f(x + 1) -1 (2, 1)

  5. (1 , 4) f(x) 0 (3, 1) Transformations of Points • f(x – a) Imagine a function where y = f(x), which has a root at 0, and points (1 , 4) and (3 , 1) lie on the curve: (2 , 4) f(x - 1) 1 (4, 1)

  6. (1 , 4) f(x) 0 (3, 1) Transformations of Points • f(x) - a Imagine a function where y = f(x), which has a root at 0, and points (1 , 4) and (3 , 1) lie on the curve: f(x) - 4 (1 , 0) -4 (3, -3)

  7. (1 , 4) f(x) 0 (3, 1) Transformations of Points • nf(x) Imagine a function where y = f(x), which has a root at 0, and points (1 , 4) and (3 , 1) lie on the curve: (1 , 8) 2f(x) (3, 2) 0

  8. (1 , 4) f(x) 0 (3, 1) Transformations of Points • f(nx) Imagine a function where y = f(x), which has a root at 0, and points (1 , 4) and (3 , 1) lie on the curve: (0.5 , 4) f(2x) (1.5, 1) 0

  9. (1 , 4) f(x) 0 (3, 1) Transformations of Points • -f(x) Imagine a function where y = f(x), which has a root at 0, and points (1 , 4) and (3 , 1) lie on the curve: -f(x) 0 (3, -1) (1 , -4)

  10. (1 , 4) f(x) 0 (3, 1) Transformations of Points • f(-x) Imagine a function where y = f(x), which has a root at 0, and points (1 , 4) and (3 , 1) lie on the curve: (-1 , 4) f(-x) 0 (-3, 1)

  11. Quadratic Functions • When examining a quadratic for the transformations from y = x2 • Complete the square to get the quadratic into the form y = n(x - a)2 + b • Think about the series of transformations from that…

  12. Quadratics Example… • y = -4x2 + 8x + 3 • = -4(x2 – 2x) + 3 [ Factorise the 4 ] • = -4[ (x – 1)2-1 ] + 3 [Complete the sq] • = -4(x – 1)2 + 7 [ (-4 x -1) + 3 = 7]

  13. Quadratic Example • y = -4x2 + 8x + 3 = -4(x – 1)2 + 7 y = x2 y = (x – 1)2 y = 4(x – 1)2

  14. Quadratic Example • y = -4x2 + 8x + 3 = -4(x – 1)2 + 7 y = 4(x – 1)2 y = - 4(x – 1)2 y = - 4(x – 1)2 + 7 Line of symmetry: x = 1, max at +7, intercept: +3

More Related