Influence of a transformation of the equation of a function on the graph

• f(x) and f(x-a)
  
• f(x) and f(x) + a
  
• Translation of the graph of y=f(x) by a vector v
  
• f(x) and f(a.x)
  
• Composing transformations
  
  • Example 1
    
  • Example 2
    
  • Example 3
    
  • Example 4
    
  • Example 5
    
• Solved Problems
  

f(x) and f(x-a)

 
y  = 3 x2 + 4 x + 5    and

y  = 3 (x-3)2 + 4 (x-3) + 5

The image of a by the first function is

3. a² + 4.a + 5

If we want the same image for the second function, we have to replace x by a+3. Then we have the image

3. ((a+3)-3)² + 4.((a+3)-3) + 5 = 3. a² + 4.a + 5

Generalizing: the image of x by the first function is the same as the image of x+3 by the second function.

If we write the first function as f(x), then the second function is f(x-3). This means that the graph of y=f(x-3) arises when we move the graph of y = f(x) three units to the right.

Exercise : plot the two functions and notice the translation.

The graph of y=f(x-a) arises when we move the graph of y = f(x) exactly a units to the right.

f(x) and f(x) + a

For all x, we see that the image of f(x) + a is exactly a units more than the image of f(x).

The graph of y=f(x) + a arises when we move the graph of y = f(x) exactly a units upwards.

Translation of the graph of y=f(x) by a vector v

Conclusion:

Is we submit the graph of y= f(x) to a translation by the vector v(a,b), then the image is the function with equation y = f(x-a) + b.

f(x) and f(a.x)

To realize what this means, we can take an example:

 
y  =  x2 - x   en

y  = (3x)2 - (3x)

a² - a

If we want the same image for the second function, we have to start with x = a/3. Then we have the image

(3. a/3 )² - (3. a/3) = a² - a

Generalizing: the image of x by the first function is the same as the image of x/3 by the second function.

This means that the graph of y=f(3x) arises when we compress the graph of y = f(x) towards the y-axis with a factor 3.

The graph of y=f(ax) arises when we compress the graph of y = f(x) towards the y-axis with a factor a.

Note: compress the graph of y = f(x) towards the y-axis with a factor 1/4 is the same as stretching the graph away from the y-axis with factor 4.

Exercise: plot y = x² and y = (x/4)².

Composing transformations

Example 1

• First, we move the graph 2 units to the right.
  The equation becomes y = sin(x - 2)
• Now, we multiply all images with the factor 3.
  The equation becomes y = 3 sin(x - 2)
• Finally, we translate the graph 5 units upwards.
  The equation becomes y = 3 sin(x - 2) + 5

Example 2

• First, we move the graph 1 units to the left.
  The equation becomes y = 3( (x+1) -1 )² - 4 <=> y = 3 x² - 4
• Now, we translate the graph 4 units upwards.
  The equation becomes y = 3 x²
• Now, we multiply all images with factor 1/3.
  The equation becomes y = x²

Example 3

• First, we compress the graph towards the y-axis with a factor 4.
  The equation becomes y = (4x)² - 4x <=> y = 16 x² - 4x
• Now, we translate the graph 2 units to the left.
  The equation becomes y = 16 (x+2)² - 4(x+2) <=> y = 16x² + 60x + 56

Example 4

• First, we translate the graph 4 units downwards.
  The equation becomes y = 5 cos(2x-6) <=> y = 5 cos(2 (x-3))
• Now, we translate the graph 3 units to the left.
  The equation becomes y = 5 cos(2x)
• We multiply all images with factor 1/5.
  The equation becomes y = cos(2x)
• Finally, we stretch the graph away from the y-axis with factor 2.
  The equation becomes y = cos(x).

Example 5

• We compress the graph of y = f(x) towards the y-axis with factor 2.
  The equation becomes y = cos(2x).
• We multiply all images with factor 5.
  The equation becomes y = 5 cos(2x)
• Now, we translate the graph 3 units to the right.
  The equation becomes y = 5.cos(2(x-3))
• Finally, we translate the graph 4 units upwards.
  The equation becomes y = 5.cos(2x -6) + 4.

Solved Problems

• 
  

  We translate the parabola with equation y = 2x2 to the right such that the vertex is on the line y = 2x - 5 Find the equation of the translated parabola.

  
  

  The intersection of the line with the x-axis is point (5/2 , 0).
  We have to translate the parabola to the right by a distance 5/2.
  The equation of translated parabola is

   
    y = (x - 5/2)2
  <=>
    y = x2 - 5x + 25/4
  

• 
  

  We translate the parabola y = x2 upwards until the line y = x is tangent to the parabola. Find the equation of the translated parabola.

  
  

  We translate the parabola a units upwards. Then , the equation is y = x² +a.
  

   
    The line  y = x is tangent to the parabola y = x2 +a
  <=>
    There is only one intersection point of   y = x and  y = x2 +a
  <=>
    The following system has two equal solutions
    / y = x
    \ y = x2 +a
  <=>
     x2 - x + a = 0  has two equal solutions
  <=>
     1 - 4a = 0
  <=>
     a = 1/4
  

  The translated parabola is y = x² + 1/4

• 
  

  The parabola y = x2 is compressed towards the y-axis. This compressed parabola intersects the line y=x in point S(1/4,1/4). Find the equation of the transformed parabola.

  
  

  The parabola y = x² is compressed towards the y-axis with a factor a.
  Then the equation of the parabola is y = (ax)².
  S is on the parabola if and only if 1/4 = a² .(1/16). So, a =2.
  The transformed parabola has an equation y = 4 x²

• 
  

  The point S(1,1) is a point of the parabola y=x2. We stretch the parabola away from the y-axis until the distance from O(0,0) to S is doubled relative to the original distance. Find the equation of the stretched parabola.

  
  

  We stretch the parabola away from the y-axis with factor a. Then the equation of the parabola is y=(x/a)² or also y=x²/a².
  Now, point S has (a,1) as coordinates and the distance |OS|² is 1 + a².
  The original value of |OS|² was 2. Thus, in the stretched position, |OS|² must be 8.
  1 + a² = 8 <=> a² = 7 . The stretched parabola has y = x²/7 as equation.

• 
  

  We compress the graph of y = 3 sin(2 x - 1) towards the y-axis with a suitable factor a such that the period of the new graph is equal to 1. Find the value of a and the equation of the new graph.

  
  

  After the compression with factor a, the new equation is y = 3 sin(2a x - 1).
  The period of this function is (2 pi)/(2 a).
  The period is 1 if and only if a = pi.
  Then the equation is y = 3 sin(2pi x - 1).

• 
  

  We translate the graph of y = 2 cos2(2x) downwards by a suitable distance v such that the translated graph is symmetric relative to the x-axis. Find v and the equation of the translated graph.

  
  

  By the Carnot-formulas, we know that 2 cos²(2x) = 1 + cos(4x).
  Then, the given equation is y = cos(4x) + 1.
  We translate the graph 1 unit downwards.
  The equation of the translated graph is y = cos(4x) and v = 1.

• 
  

  We translate the graph of y = 3 cos(2 pi x + 4) to the left by a positive distance a. Find the minimum value for a such that the origin O(0,0) is on the translated graph.

  
  

  The equation of the translated graph is

   
      y = 3 cos(2 pi (x+a) + 4).
  <=>
      y = 3 cos(2 pi x + 4 + 2 pi a)
  
      O(0,0) is on the translated graph.
  <=>
         0 = 3 cos(4 + 2 pi a)
  <=>
       cos(4 + 2 pi a) = 0
  <=>
      4 + 2 pi a = pi/2 + 2 k pi   of 4 + 2 pi a = - pi/2 + 2 k pi
  <=>
      ...
  <=>
      a = -0.3866 + k              of   a = -0.8866 + k
  

  The minimum positive value for a is 0.11338
  

  Tip: Check this result with a plotter

• 
  

  We translate the graph of y = cos(x+2) + cos(x-2) upwards by a positive distance b such that the translated graph is tangent to the x-axis. Find the value of b.

  
  

  The translated graph has as equation

   
      y = cos(x+2) + cos(x-2) + b
  <=>
      y = 2 cos(x) cos(2) + b
  <=>
      y = 2 cos(2) cos(x) + b
  

  Since 2 cos(2) is negative, the value of b is -2 cos(2).

  Tip: Check this result with a plotter