240 likes | 509 Views
Warm Up Section 3.3 (1). Solve: 2 x – 3 = 12 (2). Solve and graph: 3 x + 1 ≤ 7 (3). Solve and graph: 2 – x > 9 (4). {(0, 3), (1, -4), (5, 6), (4, 6)} State domain, range. Is relation a function? If function, is it one-to-one?. Solutions for Warm Up Section 3.3
E N D
Warm Up Section 3.3 (1). Solve: 2x – 3 = 12 (2). Solve and graph: 3x + 1 ≤ 7 (3). Solve and graph: 2 – x > 9 (4). {(0, 3), (1, -4), (5, 6), (4, 6)} State domain, range. Is relation a function? If function, is it one-to-one?
Solutions for Warm Up Section 3.3 (1). Solve: 2x – 3 = 12 (2). Solve and graph: 3x + 1 ≤ 7 (3). Solve and graph: 2 – x > 9 (4). {(0, 3), (1, -4), (5, 6), (4, 6)} State domain, range. Is relation a function? If function, is it one-to-one? 15/2 or -9/2 -8/3 2 -7 11 D = 0, 1, 5, 4 R = 3, -4, 6 Function Not one-to-one
3.2 Homework D: {0, 1, 2, 3, 4}, R: { 3, 1, 2, 4}, F: Yes; 1-to-1: No D: {-2, -1, 0, 1}, R: {-3, -1, 1, 3, 5}, F: No; 1-to-1: No D: {1, 2, 3}, R: {1, 2, 3}, F: No; 1-to-1: No D: {-2, -1, 4, 0, 2}, R: {1}, F: Yes, 1-to-1: No D: {1, 0, 5, 2}, R: {4, 2, -3, 0}, F: No; 1-to-1: No D: {x: x ≤ 7 }, R: {y: y ≤ 4}, F: Yes, 1-1:No F: Yes, 1-to-1: No 8. F: Yes, 1-to-1: No F: Yes, 1-to-1: Yes 13. No; f(-2) = 2, f(1) = 2 14. Yes; f(2) = -3, f(-3) = 12 15. No; f(-4) = 2, f(0) = 2
10. 11. 12.
Graphing Absolute Value Functions Section 3.3 Standard: MM2A1 b Essential Question: How do I graph and describe an absolute value function?
Vocabulary: Absolute value function: f(x) = |x| Vertex of an absolute value graph: the highest or lowest point on the graph of an absolute value function Transformation: changes a graph’s size, shape, position or orientation Translation: a transformation that shifts a graph horizontally and/or vertically, but doesn’t not change its size, shape or orientation
Reflection: when a = –1, the graph y = a|x| flips over the x-axis Axis of symmetry: a vertical line that divides the graph into mirror images Zeros: values of x that make the value of f(x) = 0
Recall the characteristics and graph of y = x 0 1 2 1 2 Axis of Symmetry
Recall the characteristics and graph of y = x Identify each of the following: (a). Domain: ______________ (b). Range: _______________ (c). Vertex: _______________ (d). Axis of symmetry: ______ (e). Opens: _______________ (f). Max or min: ___________ (g). x-intercept: ____________ (h). y-intercept: ___________ (i). Extrema: ______________ (j). Increasing: ____________ (k). Decreasing: ____________ (l). Rate of change: ________ All reals y≥ 0 (0, 0) x = 0 Upward Min (0, 0) (0, 0) Min value = 0 x≥ 0 x≤ 0 Left: -1; Right: 1
Graph y = |x + 3| + 2. Recall: By adding 3 to x, we are shifting the parent graph 3 units to the left. By adding 2 to y, we are then shifting the graph 2 units upward. Hence, the vertex is now at (-3, 2) and the basic shape is the same.
1. Graph y = |x + 3| + 2. Identify each of the following: (a). Domain: ______________ (b). Range: _______________ (c). Vertex: _______________ (d). Axis of symmetry: ______ (e). Opens: _______________ (f). Max or min: ___________ (g). x-intercept: ____________ (h). y-intercept: ___________ (i). Extrema: ______________ (j). Increasing: ____________ (k). Decreasing: ____________ (l). Rate of change: ________ All reals y≥ 2 (-3, 2) x = -3 Upward Min None (0, 5) Min value = 2 x≥ -3 x≤ -3 Left: -1; Right: 1
2. Graph y = -| x– 4 | + 3. Vertex: (4, 3) Opens Down Slope is 1 and -1.
2. Graph y = -| x– 4 | + 3. Identify each of the following: (a). Domain: ______________ (b). Range: _______________ (c). Vertex: _______________ (d). Axis of symmetry: ______ (e). Opens: _______________ (f). Max or min: ___________ (g). x-intercept: ____________ (h). y-intercept: ___________ (i). Extrema: ______________ (j). Increasing: ____________ (k). Decreasing: __________ (l). Rate of change: ________ All reals y≤ 3 (4, 3) x = 4 Downward Max (1, 0) and (7, 0) (0, -1) Max value = 3 x≤ 4 x≥ 4 Left: 1; Right: -1
Using the graph of y = -| x – 4 | + 3, solve each equation or inequality: (a). -| x – 4 | + 3 = 0 (b). -| x – 4 | + 3 > 0 (c). -| x – 4 | + 3 < 0 x = 1, 7 1 < x < 7 x < 1or x > 7
4. Graph Vertex: (1, -2) Opens Up Slope is ½ and -½.
4. Graph Identify each of the following: (a). Domain: ______________ (b). Range: _______________ (c). Vertex: _______________ (d). Axis of symmetry: ______ (e). Opens: _______________ (f). Max or min: ___________ (g). x-intercept: ____________ (h). y-intercept: ___________ (i). Extrema: ______________ (j). Increasing: ____________ (k). Decreasing: ____________ (l). Rate of change: ________ (-∞,∞) [-2, ∞) or y ≥ -2 (1, -2) x = 1 Upward Min (5, 0) and (-3, 0) (0, -1.5) Min value = -2 [1, ∞) or x ≥ 1 (-∞, 1] or x ≤ 1 Left: -1/2; Right: 1/2
5. Using the graph of , solve each equation or inequality: (a). (b). (c). x = -3, 5 x < -3or x > 5 -3 < x < 5
6. Graph y = -3|x + 2| – 5 Vertex: (-2, -5) Opens Down Slope is 3 and -3.
6. Graph y = -3|x + 2| – 5 Identify each of the following: (a). Domain: ______________ (b). Range: _______________ (c). Vertex: _______________ (d). Axis of symmetry: ______ (e). Opens: _______________ (f). Max or min: ___________ (g). x-intercept: ____________ (h). y-intercept: ___________ (i). Extrema: ______________ (j). Increasing: ____________ (k). Decreasing: ____________ (l). Rate of change: ________ All reals y≤ -5 (-2, -5) x = -2 Downward Max None (0, -11) Max value = -5 x≤ -2 x≥ -2 Left: 3; Right: -3
7. Write a function for the graph shown. The vertex of the graph is _______________. So the graph has the form y = a|x - ____| + _____. To determine the value of a, compute the absolute value of the slope from (1, -2) to (3, 2). (3, 2) (3, 2) 2 3 (1, -2) m = 4/-2 2 = 2 y = -2x – 3+2 An equation for the graph is _____________________
8. Write a function for the graph shown. Vertex: _______________. y = a|x - ____| + _____ (4, 5) (-2, 3) -2 3 (-2, 3) To determine the value of a, compute the absolute value of the slope from (-2, 3) to (4, 5). m = 2/6 1/3 = 1/3 An equation for the graph is _____________________
9. Without graphing, determine the intercepts for the graph of y = 5x + 12 – 45 x-intercept: Let y = 0 0 = 5x + 12 – 45 45 = 5x + 12 9 = x + 12 x + 12 = 9 x = -3 OR x + 12 = -9 x = -21 x-intercepts: (-3, 0) and (-21, 0)
9. Without graphing, determine the intercepts for the graph of y = 5x + 12 – 45 y-intercept: Let x = 0 y = 50 + 12 – 45 y = 512 – 45 y = 5(12) – 45 y = 60 – 45 y = 15 y-intercept: (0, 15)
10. Without graphing, determine the values of x for which y = 5x + 12 – 45 lies above the x-axis. 5x + 12 – 45 > 0 5x + 12 > 45 x + 12 > 9 x + 12 > 9 x > -3 OR x + 12 < -9 x < -21 x < -21 or x > -3