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Graph y = 4(x-2)² + 3 using transformations. y= 4(x- 2)2 + Use a series of transformations on the graph of y = x2 to produce the graph of y = 4(x- 2)2 +3. The graph of y = x2 is shown to the right. (1,7) (3,7) Begin by transforming y = x2 to y = 4x². For a>0, the graph of y = af(x) is a vertical stretch or compression of the graph of y = f(x). So, either a vertical stretch or compression will be used to change the graph of y = x2 to the graph of y = 4x². 2- (23) do 642 6 8 10 -2- Note that for a> 0, the graph of y = af(x) is a vertical stretch of the graph of y = f(x), if a > 1, and is a vertical compression of the graph of y = f(x), if 0 0, the graph of y = f(x ± h) is a horizontal shift of the graph of y = f(x). So, a horizontal shift should be used to change the graph of y = 4x2 to the graph of y = 4(x - 2)?. Note that for h> 0, the graph of y = f(x – h) is the graph of f(x) shifted horizontally right h units, and the graph of y = f(x +h) is the graph of f(x) shifted horizontally left h units. Hence, the graph of y = 4(x - 2)2 is the

Graph y = 4(x-2)² + 3 using transformations. y= 4(x- 2)2 + Use a series of transformations on the graph of y = x2 to produce the graph of y = 4(x- 2)2 +3. The graph of y = x2 is shown to the right. (1,7) (3,7) Begin by transforming y = x2 to y = 4x². For a>0, the graph of y = af(x) is a vertical stretch or compression of the graph of y = f(x). So, either a vertical stretch or compression will be used to change the graph of y = x2 to the graph of y = 4x². 2- (23) do 642 6 8 10 -2- Note that for a> 0, the graph of y = af(x) is a vertical stretch of the graph of y = f(x), if a > 1, and is a vertical compression of the graph of y = f(x), if 0 0, the graph of y = f(x ± h) is a horizontal shift of the graph of y = f(x). So, a horizontal shift should be used to change the graph of y = 4x2 to the graph of y = 4(x - 2)?. Note that for h> 0, the graph of y = f(x – h) is the graph of f(x) shifted horizontally right h units, and the graph of y = f(x +h) is the graph of f(x) shifted horizontally left h units. Hence, the graph of y = 4(x - 2)2 is the

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you please help me drawing a graph. I am attaching a picture of an example here with questions. 

Graph y = 3(x +2)2 – 1 using transformations. Ay 10- Which of the following transformations is required to graph the given function? 8- 6- Vertically stretch the graph of y = x2 by a factor of 3, shift it 2 units to the left, and 1 unit down. B. Vertically stretch the graph of y = x2 by a factor of 3, shift it 2 units to the right, and 1 unit down. 4- O C. Vertically stretch the graph of y = x2 by a factor of 3, shift it 2 units to the right, and 1 unit up. 2- O D. Vertically stretch the graph of y = x2 by a factor of 3, shift it 1 unit to the left, and 2 units down. 10 -8 -4 8. 10 Use the graphing tool to graph the equation. -2- -4- Click to enlarge graph -6- -8-
Transcribed Image Text:Graph y = 3(x +2)2 – 1 using transformations. Ay 10- Which of the following transformations is required to graph the given function? 8- 6- Vertically stretch the graph of y = x2 by a factor of 3, shift it 2 units to the left, and 1 unit down. B. Vertically stretch the graph of y = x2 by a factor of 3, shift it 2 units to the right, and 1 unit down. 4- O C. Vertically stretch the graph of y = x2 by a factor of 3, shift it 2 units to the right, and 1 unit up. 2- O D. Vertically stretch the graph of y = x2 by a factor of 3, shift it 1 unit to the left, and 2 units down. 10 -8 -4 8. 10 Use the graphing tool to graph the equation. -2- -4- Click to enlarge graph -6- -8-
Graph y = 4(x - 2)2 + 3 using transformations. 10어 y = 4(x – 2)² +3 8- Use a series of transformations on the graph of y=x2 to produce the graph of y = 4(x- 2)² + 3. The graph of y = x2 is shown to the right. (1,7) 4- (3,7) Begin by transforming y = x2 to y = 4x2. For a> 0, the graph of y = af(x) is a vertical stretch or compression of the graph of y= f(x). So, either a vertical stretch or compression will be used to change the graph of y = x2 to the graph of y = 4x2. 러 (2,3) -10-8 -6 -2 -2- 10 -4 -4- Note that for a>0, the graph of y = af(x) is a vertical stretch of the graph of y = f(x), if a> 1, and is a vertical compression of the graph of y = f(x), if 0<a<1. Hence, the graph of y = 4x² is a vertical stretch by a factor of 4 of the graph of y = x2. -8- 10 In the function y = 4x², the output of the function is being multiplied by 4. When an output of a function f(x) is multiplied by a positive number a, the graph of the new function y = af(x) is obtained by multiplying each y-coordinate on the graph of y = f(x) by a. Consider the three points on the graph of y = x2. Find the coordinates of the corresponding points that lie on the graph of the vertical stretch of the graph of y = x2 by a factor of 4. Multiply the y-coordinate by 4 and keep the x-coordinate the same. Points that lie on the graph of y = x2 Corresponding points that lie on the graph of y = 4x? (- 1,1) (0,0) (1,1) (- 1,4) (0,0) (1,4) Using the points found in the previous step, vertically stretch the graph of y = x2 by a factor of 4 to sketch the graph of y = 4x². The correct graph is shown to the right. Use a second transformation to change y = 4x2 to y = 4(x – 2)2. For h>0, the graph of y = f(x +h) is a horizontal shift of the graph of y = f(x). So, a horizontal shift should be used to change the graph of y = 4x2 to the graph of y = 4(x – 2)². Note that for h > 0, the graph of y = f(x – h) is the graph of f(x) shifted horizontally right h units, and the graph of y = f(x + h) is the graph of f(x) shifted horizontally left h units. Hence, the graph of y = 4(x- 2) is the
Transcribed Image Text:Graph y = 4(x - 2)2 + 3 using transformations. 10어 y = 4(x – 2)² +3 8- Use a series of transformations on the graph of y=x2 to produce the graph of y = 4(x- 2)² + 3. The graph of y = x2 is shown to the right. (1,7) 4- (3,7) Begin by transforming y = x2 to y = 4x2. For a> 0, the graph of y = af(x) is a vertical stretch or compression of the graph of y= f(x). So, either a vertical stretch or compression will be used to change the graph of y = x2 to the graph of y = 4x2. 러 (2,3) -10-8 -6 -2 -2- 10 -4 -4- Note that for a>0, the graph of y = af(x) is a vertical stretch of the graph of y = f(x), if a> 1, and is a vertical compression of the graph of y = f(x), if 0<a<1. Hence, the graph of y = 4x² is a vertical stretch by a factor of 4 of the graph of y = x2. -8- 10 In the function y = 4x², the output of the function is being multiplied by 4. When an output of a function f(x) is multiplied by a positive number a, the graph of the new function y = af(x) is obtained by multiplying each y-coordinate on the graph of y = f(x) by a. Consider the three points on the graph of y = x2. Find the coordinates of the corresponding points that lie on the graph of the vertical stretch of the graph of y = x2 by a factor of 4. Multiply the y-coordinate by 4 and keep the x-coordinate the same. Points that lie on the graph of y = x2 Corresponding points that lie on the graph of y = 4x? (- 1,1) (0,0) (1,1) (- 1,4) (0,0) (1,4) Using the points found in the previous step, vertically stretch the graph of y = x2 by a factor of 4 to sketch the graph of y = 4x². The correct graph is shown to the right. Use a second transformation to change y = 4x2 to y = 4(x – 2)2. For h>0, the graph of y = f(x +h) is a horizontal shift of the graph of y = f(x). So, a horizontal shift should be used to change the graph of y = 4x2 to the graph of y = 4(x – 2)². Note that for h > 0, the graph of y = f(x – h) is the graph of f(x) shifted horizontally right h units, and the graph of y = f(x + h) is the graph of f(x) shifted horizontally left h units. Hence, the graph of y = 4(x- 2) is the
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