Batman. Part 1 and 2. Sam wants to recreate the Batman symbol using graphs. Describe fully the brown, orange and blue graphs.

Transcription

1 Batman Part 1 and 2 Sam wants to recreate the Batman symbol using graphs. Describe fully the brown, orange and blue graphs. Sketch and describe the following graphs teal: y = sinx 14 starting at x = -15 and finishing at x = 15 purple: y = 0.28(x 6) starting at x = 3 and finishing at x = 10 green: y = 2 x 13 starting at x = -3 and finishing at x=3

2 Part 3 Sam then decides to also create a picture of a bat using her new knowledge of graphs. Your task is to sketch, provide and describe at least one equation that models any part of the bat.

3 Part 4: Sam wants a generalised model of the top border of the Batman mask. Part of the shape is shown by the solid line in the diagram to the right. y The complete design may be needed to cross X masks of varying widths. The highest point will be 2 centimetres above the x-axis. The border is symmetrical horizontally. The total width of the border will be w centimetres. Generalise your models in to meet these requirements. For each model: Give the equation of any function(s) used in your model. Discuss the limitations for each of the functions used in your model.

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5 Brown: Sam we are creating an equation for the top curved left-hand side of the wing Parabola, vertex form y = x 2 Vertex (0,0) Transformed graph vertex at (-6, 3.5) Axis of symmetry x = -6 y = a(x ± b) 2 + c Explanation of a, b, c, d, b = horizontal shift left 6 b = 6 as x + 6 = 0 c = vertical shift up 3. 5 c = 3. 5 a = stretch To calculate a substitute in a point on the graph (-3, 6) so y = a(x + 6) = a( 3 + 6) = 9a 5 18 = a or 0.28 Equation of graph Domain y = 5 18 (x + 6) x 3

6 Orange: Sam we are creating an equation for top of left ear Absolute value y = x Vertex (0,0) Transformed graph vertex at (-2, 10) Axis of symmetry x=-2 y = a x ± b ± c Explanation of a, b, c, d, b = horizontal shift left 2 b = 2 as x + 2 = 0 c = vertical shift up 10 c = 10 To graph has been reflected in the x-axis and therefore the gradient will be negative a = gradient To calculate gradient use two points on the graph (-3, 9) and (-2, 10) so gradient = 10 9 = 1 therefore, it a = Equation of graph Domain y = 1 x x 1

7 Blue: Sam we are creating an equation for the bottom curved left-hand side of the wing Parabola, x intercept form y = x 2 Vertex (0,0) Transformed graph has an axis of symmetry x=- 4.5 ( 6 + 3) 2 = 4. 5 y = a(x ± b)(x ± c) ± d Explanation of a, b, c, d, b and c are x intercepts before transformed vertically ( 6, 0) ( 3, 0) x + 6 = 0 and x + 3 = 0 b = 6 and c = 3 d = vertical shift down 7 c = 7 a = stretch will be - negative as reflected in the x-axis (flipped over) To calculate a substitute in a point on the graph (-3.5,-5.5) so 5. 5 = a( )( ) = a( )( ) 1. 5 = a(2. 5)( 0. 5) 1.25 = a Equation of graph Domain y = 1.25(x + 6)(x + 3) 7 6 x 3

8 Teal: Sam we are going to discuss how to model/sketch the wave at the bottom of the sketch y = sinx 14 starting at x = -15 and finishing at x = 15 Sine trig graph y = sin (x) Symmetric about the origin Periodic of 2π Domain is (-, ) Range [-1, 1] y = Asin(Bx C) + D Explanation of A, B, C, D, A = amplitude (Half the total height of the wave) A = 1 B = period this has 1 for B therefore one period has a length of 2π D = vertical shift down 14 D = 14 C = phase shift there is no phase shift or horizontal shift y = sin (x) 14 Domain 15 x 15

9 Purple: Sam we are going to discuss how to model/sketch the equation for the top curved righthand side of the wing y = 0.28(x 6) starting at x = 3 and finishing at x = 10 Parabola, vertex form y = x 2 Vertex (0,0) Transformed graph reflects the brown graph through x=0 Vertex (6, 3.5) Axis of symmetry x = 6 for purple graph y = a(x ± b) 2 + c Equation of graph Domain y = 5 18 (x 6) x 10

10 Green: Sam we are going to discuss how to model/sketch the equation for the tail of the symbol y = 2 x 13 starting at x = -3 and finishing at x=3 Absolute value y = x Vertex (0,0) Transformed graph has vertex (0, -13) y = a x ± b ± c Explanation of a, b, c, d, b = no horizontal shift so B=0 c = vertical shift down 13 c = 13 a = gradient To gradient is 2 so a = 2 Two units up for every 1 unit across 1 Domain 3 x 3 Two possible options for the Bat. Write-up also needs to be included.

11 The Mask Possible solutions. All have domain 0 x w 2 1. Square root with vertex at (0,0), y = x. Substitute in the point ( w 2,2) giving 2 = k w 2 so k = 2. The equation is y = 2 x w 2 w 2 The square root graph, with a line of reflection being x = w. This could give a curved shape 2 to reflect the brow of the mask but the point of intersection would not be smooth. This graph has a steeper incline for values closer to x=0 which may not model the smooth brow. The reflected model would be y = 2 w x w 2 2. Sine curve with amplitude of 2 at x = w 2 and a period of π w. y = k sin(x) giving y = 2 sin(π w x) The sine curve is smooth and curved at the point ( w,2). It is steeper than the square root 2 graph for the lower values of the domain. The ends may be too steep to model the brow. It is smooth at x= w 2 3. Log curve. y = k log x has shifted one unit to the left, at the x-intercept point (0,0), so y = k log(x + 1), substitute in point on the curve ( w 2,2) 2 = k log (w 2 + 1) so k = 2 log ( w 2 +1). Therefore y = 2 log(x+1) log ( w 2 +1) The log curve is the steepest of all the equations. As x increases y increases at the greatest rate. The line of reflection is also x= w. The point of intersection is not smooth. 2 2 = k log ( w 2 + w + 1) so k = 2 log ( w + w + 1) 2 The reflected model would be y = 2 log( x+w+1) log ( w 2 +w+1)