Back to Baker Street... Lesson 3A Assignment Complete the Back to Baker Street assignment. Refer to the Tips From Scotland Yard by clicking on the link to assist you in completing the assignment. Call or e-mail your teacher if you would like additional help. Name: <> Date: <> Score: < /32 = % > MultipleChoice Items 1. <> 2. <> Numeric Response 1 < 3. <> > 4. <> Numeric Response 2 < 5. <> > 6. <> Numeric Response 3 < 7. <> > Written Response 1. <> <> <> 2. <> <> 3. <> <> <> Back to Baker Street... Lesson 3B Assignment Complete the Back to Baker Street assignment. Refer to the Tips From Scotland Yard by clicking on the link to assist you in completing the assignment. Call or e-mail your teacher if you would like additional help. Name: <> Date: <> Score: < /36 = %> MultipleChoice Items 1. <> 2. <> Numeric Response 1 < 3. <> > 4. <> Numeric Response 2 < 5. <> 6. <> > Numeric Response 3 < Written Response 1. <> <> > <> 2. <> <> <> Back to Baker Street... Lesson 4A Assignment Complete the Back to Baker Street assignment. Refer to the Tips From Scotland Yard by clicking on the link to assist you in completing the assignment. Call or e-mail your teacher if you would like additional help. Name: <> Date: <> Score: < /25 = % > MultipleChoice Items 1. <> 2. <> 3. <> Numeric Response 1 < > Numeric Response 2 < 4. <> > 5. <> 6. <> Numeric Response 3 < Written Response 1. <> 2. <> > Back to Baker Street... Lesson 4B Assignment Complete the Back to Baker Street assignment. Refer to the Tips From Scotland Yard by clicking on the link to assist you in completing the assignment. Call or e-mail your teacher if you would like additional help. Name: <> Date: <> Score: < /33 = %> MultipleChoice Items 1. <> 2. <> 3. <> Numeric Response 1 < > 4. <> Numeric Response 2 < 5. <> > 6. <> Written Response 1. <> 2. Back to Baker Street... Lesson 5A Assignment Complete the Back to Baker Street assignment. Refer to the Tips From Scotland Yard by clicking on the link to assist you in completing the assignment. Call or e-mail your teacher if you would like additional help. Name: <> Date: <> Score: < /35 = % > Multiple-Choice Items 1. <> 2. <> Numeric Response 1 < 3. <> > 4. <> Numeric Response 2 < 5. <> > 6. <> 7. <> Numeric Response 3 < > Written Response 1. <> <> <> 2. <> <> <> 3. <> <> Back to Baker Street... Lesson 3A Assignment Complete the Back to Baker Street assignment. Refer to the Tips From Scotland Yard by clicking on the link to assist you in completing the assignment. Call or e-mail your teacher if you would like additional help. Name: <> Date: <> Score: < /32 = % > Multiple-Choice Items 1. <> Answer A Explanation= In Second function put x=z/2, it become y=f(2*z/2)=f(z) f(z)=f(x) since z and x are both variables this implies that the second function achieves the same value as that of first function but at points half way. 2. <> Answer C... Explanation the negative factor implies reflection about x axis ...while the factor 3 imples vertical stretch . Numeric Response 1 < 6 > Explanation: the new function is obtained by shrinking by factor of 4 in x direction and multiplying the by 2 in y direction the corresoponding point may be obtained by multiplying the x coordinate by factor and the y coordinate by the factor of 2 thus point is (1/4,6) 3. <> Answer B...to find coordinate pair ..put p-3=4...p=7 now put this value in latter function you will get y =-11...the coordinate point pair is (p, y) =(7 , -11) 4. <> Answers is D.........go step by step ....stretching by factor of means f(x) becomes f(2x) ......Stretching vertically means f(2x) becomes 3f(2x)..reflection in x axis makes the function -3f(2x)...lastly, translation 4 units left means replace x by x+4...so function becomes .. -3f(2(x+4)) =-3f(2x+8) Numeric Response 2 < 3 > Explanation: TO find vertex of second function ...make the argument of function zero that is ....... 5(x-3)=0 this gives x=3 ....to find y coordinate pu this value in the second function ...whilch gives 7..thaus the vertex is at (3, 7) 5. <> B is incorrect ...not all functions have a inverse...for example take sin(x) over the interval of 0 to pi... its inverse cannot be defined. Please note that is why sin-1(x) is defined on that -pi/2 to pi /2 6. <> Answer C To invert the function replace f(x) by x and also replace x by y.. x=y2 +9 rearranging this for y y =_+ sqrt(x-9) now since the domain of original function is negative (x<0)..the range of inverted function should also be negative .t...thus neglecting the posive sign... y =- sqrt(x-9) Numeric Response 3 < 3 > Explanation ..x and y intercepts interchange on inversion...y intercept of 3 becomes x inercept 7. <> Answer D Translation changes only the position of graph but not the shape Written Response 1. <> Vertex of original function is (2,0) to find vertex of transformed function put x+5=2 which gives X=-3 and put it in second function to get Y=-4 therefore (-3, -4) is vertex of <> To get corresponding point Subtract 5 and 4 from x and y coordinates respectively of original point Answer is (1,12) <> To get original point add 5 and 4 to x and y coordinates of corresponding point respectively Answer is (-4,4) 2. <>domain (-3 to 4) range (0 to 4) <> Domain and range remains the same domain (-3 to 4) range (0 to 4) 3. <> <> <> \f