Graphical Methods Booklet

Transcription

1 Graphical Methods Booklet This document outlines the topic of work and the requirements of students working at New Zealand Curriculum level 7.

2 Parabola, vertex form y = x 2 Vertex (0,0) Axis of symmetry x=0 y = a(x ± b) 2 + c b = horizontal shift right 2 b = 2 as x + 2 = 0 c = vertical shift up 4 c = 4 a = stretch - negative as reflected in the x-axis (flipped over) To calculate a substitute in a point on the graph (0,1) so y = a(x 2) = a(0 2) = 4a 3 = a 4 y = 3 4 (x 2) x 5

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4 Parabola, x intercept form y = x 2 Vertex (0,0) Axis of symmetry x=0 y = a(x ± b)(x ± c) b and c are x intercepts ( 4, 0) (1, 0) x 4 = 0 and x + 1 = 0 b = 4 and c = 1 a = stretch - negative as reflected in the x-axis (flipped over) To calculate a substitute in a point on the graph (0,6) so 6 = a(0 + 4)(0 1) 6 = 4a 6 = 3 = a 4 2 y = 3 (x + 4)(x 1) 5 x 2 2

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6 Absolute value y = x Vertex (0,0) Axis of symmetry x=0 y = a x ± b ± c b = horizontal shift right 3 b = 3 as x + 3 = 0 c = vertical shift up 2 c = 2 a = gradient To calculate gradient use two points on the graph (3, -2) and (5, 0.5) so a = = y = 1.25 x x 5

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8 Square root y = x Absolute minimum at (0,0) Increasing function Positive function or begins at origin and rises in a concave downward way y = a (x ± b) ± c b = horizontal shift left 4 b = 4 as x 4 = 0 c = vertical shift up 3 c = 3 a = stretch - substitute (-3, 5) y = a (x + 4) = a ( 3 + 4) = a 1 2 = a y = 2 (x + 4) x 2

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10 Hyperbola y = 1 x Asymptotes, horizontal y=0, vertical x=0 Point symmetry about (0, 0) y = a (x±b) ± c b = horizontal shift right 3 b = 3 as x 3 = 0 c = vertical shift up 4 c = 4 a = stretch - substitute (4, 6) y = a x = a = a y = x 8 x 3

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12 Logarithm y = log x Asymptote at x=0 Goes through (1, 0) Increasing when a > 1 y = a log(x ± b) ± c b = horizontal shift right 2 b = 2 as x 2 = 0 c = vertical shift (across 1 to the right from asymptote graph is on x axis) c = 0 a = stretch - substitute (4, 1) y = a log(x 2) 1 = a log(4 2) a = 3.32 y = 3.32 log(x 2) 2 x 7

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14 Cubic y = x 3 Point symmetry Intersects axis at (0, 0) Slope not constant, positive (0, ) negative (-, 0) y = a(x ± b) 3 ± c b = horizontal shift right 4 b = 4 as x + 4 = 0 c = vertical shift down 3 c = 3 a = stretch To calculate a substitute in a point on the graph (2,-5) so y = a(x 4) = a(2 4) = a = a y = 1 4 (x 4)3 3 1 x 7

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16 Cubic, x intercept form y = x 3 Point symmetry Intersects axis at (0, 0) Slope not constant, positive (0, ) negative (-, 0) y = a(x ± b)(x ± c)(x ± d) b, c and d are x intercepts ( 1, 0) (2, 0) (4, 0) b = 1, c = 2 and d = 4 a = stretch To calculate a substitute in a point on the graph (1,2) so 3 = a(1 + 1)(1 2)(1 4) 2 = 6a 1 3 = a y = 1 (x + 1)(x 2)(x 4) 2 x 5 3

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18 Exponential y = a x Horizontal asymptote at y=0 y intercept (0, 1) Increasing y = a (x+b) ± c b = horizontal shift right 4 b = 4 as x + 4 = 0 c = vertical shift up 1 c = 1 a = stretch - substitute (5, 3) y = a (x 4) = a (5 4) + 1 a = 2 y = 2 (x 4) x 7

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20 Graphs on Casio fx-9750gii with domains The following shows how to graph y = x 2 with a domain of [ 2 x 3] Square brackets are shift + and shift Using Graphics Calculator to find Scale Factor 1. Enter any 3 known co-ordinates into your STAT menu (in order) So for graph 1, co-ordinates (1,8) (4,3) (7,8) are entered as shown. (x values in list 1, y values in list 2) GRPH (F1) GRPH1(F1) CALC (F1) X^2 (F4) 2. Equation is shown in the form of y = ax 2 + bx + c where a is the scale factor Note: 5 9 = Write the equation using the general form as previous y = a(x 4) so y = 5 9 (x 4)2 + 3

21 Rainbow A symmetrical rainbow is shown below. A sketch of half a rainbow is shown on the graph below. The shape of this section of the rainbow could be modelled by any of these functions: a parabola, trigonometric function, cubic, square root or logarithm function. Sam wants to know how to model the outside edge of part of the Rainbow shown. 10m d m The half rainbow has a width of d metres and a height of 10 metres. Generalise two possible models that meet these requirements. For each model: a. Give the equation of the function you have used. b. Discuss the limitations of the model.

22 y 10 : y=(10/log(31))log(x+1) 8 : y=(10/ 30) x 6 : y=10sin(x π/60) 4 2 : y= 1/90(x 60)x x Possible solutions. All have domain 0 x 2d 1. Square root with vertex at (0,0), y = x. Substitute in the point (d,10) giving 10 = k d so k = The equation is y = d d x The square root graph, with a line of reflection being x = d. This could give a curved shape to reflect the rainbow 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 rainbow. 2. Sine curve with amplitude of 10 at x = d and a period of π. y = k sin(x) giving y = 2d 10 sin( π 2d x) The sine curve is smooth and curved at the point (d,10). It has the steepest incline of all the models for the lower values of the domain. The ends may be too steep to model the rainbow. 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 ((d,10) 10 = k log(d + 1) so k = 10 log (d+1). 10 log(x+1) Therefore y = log (d+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=g. The point of intersection is not smooth

23 Fijian Bowl A sketch of half the Fijian Bowl is shown on the graph opposite. The maximum height of the bowl is 20cm and the bowl is 30cm wide, The shape of this section of the bowl could be modelled by any of these functions: parabola, hyperbola, trig, square root or cubic function. Draw graphs of at least three different functions that could be used to model half the bowl. 15cm Give the equation of the function used in each model. Discuss the features or properties for each of the functions you have developed. Jack wants to know how to model the outside edge of the bowl shown. d cm The bowl has a height of d cm and a width of 15cm. Generalise two possible models that meet these requirements. For each model: a. Give the equation of the function you have used. b. Discuss the limitations of the model.

24 Marking Grid 2017 Name: AS91257 Apply graphical methods in solving problems Achieved Equation Form equation from graph, model 1 Description (3 features or properties) All Required Graph Draw graph from equation, model 2 Description (3 features or properties) Equation or Graph Form equation from graph OR draw graph from equation, model 3 Description (3 features or properties) Equation Merit Form equation from graph Description (explanation of how the equation is formed) All Required Graph Draw graph from equation Description (explanation of how the equation is formed) Model 1 Excellence Equation of model Reflection of model Limitation of model Refer to Marking Schedule Model 2 Equation of model Reflection of model Limitation of model