Absolute value and reciprocal functions Chapter 7

example How many triangles are in the diagram below?

7.1 – Absolute value Chapter 7

|3| |–7| Absolute value Evaluate: For a real number a, the absolute value is written as |a| and is a positive number. For example: |5| = 5 |–5| = 5 Absolute value can be used to represent the distance of a number from zero on a real-number line. Evaluate: |3| |–7|

example Evaluate the following: a) |4| – |–6| b) 5 – 3|2 – 7| c) |–2(5 – 7)2 + 6| |4| – |–6| = 4 – 6 = –2 b) 5 – 3|2 – 7| = 5 – 3|–5| = 5 – 3(5) = 5 – 15 = –10 c) Try it!

Pg. 363-367 #1, 6, 7(A,C,E), 11, 16 Independent Practice

handout Answer the questions on the “Investigating Absolute Value Functions” worksheet to the best of your ability.

7.2 – absolute value functions Chapter 7

Absolute value functions For what values of x is the function y = |x| equivalent to y = x?  when x ≥ 0 When x < 0, what is the function represented by y = |x|?  y = –x We can write this as a piecewise function:

example Consider the absolute value function y = |2x – 3|. Determine the y-intercept and the x-intercept. Sketch the graph. State the domain and range. Express as a piecewise function. The y-intercept is at x = 0. y = |2x – 3| y = |2(0) – 3| y = |–3| y = 3 The y-intercept is (0, 3). The x-intercept is at y = 0. 0 = |2x – 3| 0 = 2x – 3 x = 3/2 The x-intercept is at (3/2, 0). b) x y -1 5 3 3/2 4

example Consider the absolute value function y = |2x – 3|. Determine the y-intercept and the x-intercept. Sketch the graph. State the domain and range. Express as a piecewise function. b) x y -1 5 3 3/2 4 D: {x | x E R} R: {y | y ≥ 0, y E R} d) The equation on the right is just y = 2x – 3. What’s the one on the left? It’s just –(2x – 3)! The x-intercept is call an invariant point because it’s a part of both functions.

example Consider the absolute value function f(x) = |–x2 + 2x + 8|. Determine the y-intercept and the x-intercepts. Sketch the graph. State the domain and range. Express as a piecewise function The y-intercept is at x = 0. f(0) = |–(0)2 + 2(0) + 8| = |8| = 8 The x-intercepts are when y = 0. 0 = |–x2 + 2x + 8| 0 = –x2 + 2x + 8 0 = –(x – 4)(x + 2) x = 4 x = –2 b) What’s the vertex of the function f(x) = –x2 + 2x + 8 Use your calculator, or complete the square: f(x) = –(x2 – 2x) + 8 f(x) = –(x2 – 2x + 1 – 1 ) + 8 f(x) = –(x2 – 2x + 1) + 1 + 8 f(x) = –(x – 1)2 + 9 The vertex is (1, 9)

example Consider the absolute value function f(x) = |–x2 + 2x + 8|. Determine the y-intercept and the x-intercepts. Sketch the graph. State the domain and range. Express as a piecewise function Recall: y-intercept is (0, 8) x-intercepts are (4, 0) and (–2, 0) Vertex is (1, 9) c) D: {x | x E R} R: {y | y ≥ 0, y E R} d)

Pg. 375-379, #2, 5, 7, 10, 12-14. Independent Practice

7.3 - Absolute value equations Chapter 7

Absolute value equations When solving equations that involve absolute value equations you need to consider two cases: Case 1: The expression inside the absolute value symbol is positive or zero. Case 2: The expression inside the absolute value symbol is negative.

Example The solution is x = 10 or x = –4. Solve: |x – 3| = 7 Consider the equation as a piecewise function: Case 1: x – 3 = 7  x = 10 Case 2: –(x – 3) = 7  x – 3 = –7  x = –4 The solution is x = 10 or x = –4.

Try it Solve |6 – x| = 2

example Solve |2x – 5| = 5 – 3x What is the x-intercept of y = 2x – 5? Case 1: (x ≥ 5/2) 2x – 5 = 5 – 3x 5x = 10 x = 2 Case 2: (x < 5/2) –(2x – 5) = 5 – 3x –2x + 5 = 5 – 3x x = 0 Consider:

example Solve: |3x – 4| + 12 = 9 |3x – 4| = –3 Is there any possible way that the absolute value of something is equal to –3?  No solution.

example Solve: |x – 10| = x2 – 10x

Pg. 389-391, #4, 5, 6, 9, 11, 22, 23 Independent Practice

7.4 – reciprocal functions Chapter 7

example Sketch the graphs of y = f(x) and its reciprocal function y = 1/f(x), where f(x) = x. Examine how the functions are related. x y = x y = 1/x –5 –1/5 –2 –1/2 –1 –1/10 –10 Undef. 1/10 10 1/2 2 1

Reciprocal functions What is the vertical asymptote? An asymptote is a line whose distance from a curve approaches zero. This graph has two pieces, that both approach the vertical asymptote, which is defined by the non-permissible value of domain of the function, and a horizontal asymptote, defined by the value that is not in the range of the function. What is the vertical asymptote? What is the horizontal asymptote?

example Consider f(x) = 2x + 5. Determine its reciprocal function y = 1/f(x). Determine the equation of the vertical asymptote of the reciprocal function. Graph the function y = f(x) and its reciprocal function y = 1/f(x). a) The reciprocal function is: c) Characteristic f(x) = 2x +5 f(x)=1/(2x + 5) x-intercept/asymptotes x-intercept at x = –5/2 Asymptote at x = –5/2 Invariant points 2x + 5 = 1 x = –2 (–2, 1) (–2,1) 2x + 5 = –1 x = –3 (–3,–1) b) The vertical asymptote is always the non-permissible values of the function. 2x + 5 = 0 2x = –5 x = –5/2 Invariant points are at y = 1 and y = –1. There is a vertical asymptote at x = –5/2

example Consider f(x) = 2x + 5. Determine its reciprocal function y = 1/f(x). Determine the equation of the vertical asymptote of the reciprocal function. Graph the function y = f(x) and its reciprocal function y = 1/f(x). Characteristic f(x) = 2x +5 f(x)=1/(2x + 5) x-intercept/asymptotes x-intercept at x = –5/2 Asymptote at x = –5/2 Invariant points 2x + 5 = 1 x = –2 (–2, 1) (–2,1) 2x + 5 = –1 x = –3 (–3,–1)

example Consider f(x) = x2 – 4. What is the reciprocal function of f(x)? State the non-permissible values of x and the equation(s) of the vertical asymptote(s) of the reciprocal function. What are the x-intercepts and y-intercepts of the reciprocal function? Graph the functions. a) c) How can I find the x-intercept of the the reciprocal? Let f(x) = 0 There is no solution, so there is no x-intercept. y-intercept: What are the non-permissible values? x2 – 4 = 0 (x – 2)(x + 2) = 0 x = 2 x = –2 The vertical asymptotes are at x = ±2 y-intercept is y = –1/4

Pg. 403-408, #3, 5, 7, 8, 9, 10, 12 Independent practice