I can Shift, Reflect, and Stretch Graphs

Greatest Integer Function ||x|| is the greatest integer less than or equal to x So if f(x) = ||x|| f(3) = 3 f(4.3) = 4 f(5.9) = 5 f(-2.1) = -3 Now graph this!

Can you think of a real-life example?

Make sure that you know these basic functions and their graphs… f(x) = c constant f(x) = x identity f(x) = x2 quadratic f(x) = x3 cubic f(x) = |x| absolute value f(x) = √x square root f(x) = 1/x reciprocal f(x) = ||x|| greatest integer

Vertical and Horizontal Shifts f(x) + c c units up (after ) f(x) – c c units down (after) f(x + c) c units left (before) f(x – c) c units right (before) Ex: y = x2 + 2, y = x2 – 2, y = (x + 2)2, y = (x – 2)2

Reflections Reflection in the x-axis - f(x) (after) Reflection in the y-axis f(-x) (before) Example 1: y = |x|, y = -|x|, y = |-x| Example 2: y = x2 – 3, y = -(x2 – 3), y = (-x)2 – 3 Example 3: y = √x, y = -√x, y = √-x

Stretches c •f(x) is a vertical expansion if c > 1 is a vertical compression if 0< c < 1 Example : y = 3x2, y = ½x2 f(c●x) is a horizontal compression if c> 1 is a horizontal expansion if 0< c < 1 Example: y = (4x)2, y = (.2x)2

Putting It All Together It is very important to think about the order of the transformations… Graph these functions y = -(x + 3)2 + 1 y = 2|x - 4| y = 4 - (x – 2)3 y = 1/2√x - 5 y = 1/(x+ 1) + 2

Piecewise Functions Graph f(x) = { x if -2 ≤ x≤ 1

Determine the transformation(s) You will need to complete the square to put each quadratic in vertex form. y= x2 – 4x + 7 right 2, up 3 y = -2x2 – 4x – 2 left one, stretched by a factor of 2, reflected over x-axis

Transformations with Absolute Value |f(x)| reflects any part of f(x) below the x-axis over the x-axis f(|x|) replaces any part of the graph to the left of the y-axis with a reflection of the part of the graph to the right of the y-axis Think about y = x and y = |x|. Now do this for y = x – 3, y = |x - 3|, and y = |x|-3.

Graph y = |x2-2|. Graph y = x3 – 4x Graph y = |x2-2|. Graph y = x3 – 4x. Now graph y = | x3 – 4x | and y = |x|3 - 4|x|.

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