Functions and Transformation of Graphs

Functions

y = x2 + 3x -2 is said to be a function of x - an expression of x. This can also be written, in a bit more advanced way as f(x) = x2 + 3x -2 - it is read as you write, function of x, f(x)
So, f(x) = x2 + 3x -2
Now whatever you put in as 'x' on the left, substitutes 'x' on the right.

E.g.
  • f(2) = 22 + 3x2 -2 = 4 + 6 - 2 = 8
  • f(-2) = (-2)2 + 3x(-2) -2 = 4 - 6 - 2 = -4
  • f(0) = 02 + 3x0 -2 = 0 + 0 - 2 = -2
  • f(x+1) = (x+1)2 + 3(x+1) -2 = x2 + 2x + 1 + 3x + 3 -2 = x2 + 5x + 2
  • f(x-1) = (x-1)2 + 3(x-1) -2 = x2 - 2x + 1 + 3x - 3 -2 = x2 + x - 5
  • f(2x) = (2x)2 + 3(2x) -2 = (2x)2 + 6x -2 = 4x2 + 6x - 2
  • f(x/2) = (x/2)2 + 3(x/2) -2 = x2/4 + 3x/2 -2 = x2/4 + 3x/2 - 2

There are three major transformations:

  1. Translation
  2. Reflection
  3. Stretching

Let's apply the above transformations to the following curve.

transfrmtns

Translation

f(x) -----> f(x+1)f(x) -----> f(x-1)
transfrmtnstransfrmtns1
translation in the -ve x-axistranslation in the +ve x-axis
f(x) -----> f(x) + 2f(x) -----> f(x) - 2
transfrmtnstransfrmtns2
translation in the +ve y-axistranslation in the -ve y-axis
translation in the x-axis

 

translation in the y-axis

Reflection


f(x) -----> f(-x)f(x) -----> -f(x)
transfrmtnstransfrmtns3
reflection in the y-axisreflection in the x-axis

Stretching


f(x) -----> f(2x)f(x) -----> 2f(x)
transfrmtnstransfrmtns
stretching parallel to x-axisstretching parallel to y-axis
stretching parallel to the x-axis

 

stretching parallel to  the y-axis

Now, let's apply the transformation to some of the well-known curves.


E.g.1

Sketch f(x) = x2 + 2x -3, and find the following:

  • line of symmetry
  • minimum value
  • translations from f(x) = x2
  • solutions
transfrmtns

First of all, let's use the completing the square method:
Let x2 + 2x -3 = (x+a)2 +b
x2 + 2x -3 = x2 + 2ax + a2+ b
Make the coefficients of x2, x and the constant on either side equal.
2a = 2 => a = 1
a2+ b = -3 => b = -4
f(x) = (x+1)2 -4
Line of symmetry, x = 1
Minimum value = -4
This is a translation of f(x) = x2, 1 in the negative x-axis and 4 in the negative y-axis.
The solutions are, x = -3 and x = 1

E.g.2

transfrmtns

 

This question is based on the curve on the left; find the coordinates of A, B and C after the following transformations. The answers are on the table.

TransformationNew Coordinates
f(x+1)A( -3 , 0 )B( -1 , 3 )C( 3 , -4 )
f(x-1) + 2A( -1 , 2 )B( 1 , 5 )C( 5 , -2 )
f(2x)A( -1 , 0 )B( 0 , 3 )C( 2 , -4 )
f(x/2)A( -4 , 0 )B( 0 , 3 )C( 8 , -4 )
2f(x)A( -2 , 0 )B( 0 , 6 )C( 4 , -8 )
f(-x) + 1A( 2 , 1 )B( 0 , 4 )C( -4 , -3 )
-f(x) + 2A( -2 , 2 )B( 0 , -1 )C( 4 , 6 )

Now, in order to complement what you have just learnt, work out the following questions:

  1. Use the completing the square method to sketch f(x) = x2 + 3x -4. Then describe the transformations of f(x) = x2 to get this particular shape. Write down the minimum value and the line of symmetry too.
    • Show that the reflection in the y-axis of f(x) = x4 - 2x2 + 5 is the same as f(x).
    • Write down a function that shows f(x) = -f(-x). Hence, sketch 2f(x) and f(2x) of the same function.
  2. If f(x) = x2 + 2x, sketch the following:
    • f(-x)
    • -f(x) + 2
    • f(x+2) - 3
    • f(2-x)
    • 2f(x) + 1
    • f(2x) - 3
  3. Sketch f(x) = x2 -x - 6 and hence sketch g(x) = (x+2)(x+3) and show that f(x+1) = g(x).
  4. Sketch f(x) = 1 +2 sin(x) and solve f(x) = 1.3 for 0 < x < 360. What is the minimum value and maximum value of the function?<
  5. Sketch f(x) = 1 + cos(3x) and hence find the period of the function, maximum value and minimum value.
  6. Sketch f(x) = 3 - 2sin(3x) and hence find the period of the function, maximum value and minimum value.
  7. Sketch f(x) = 1/x and then transform as follows:
    • f(-x) + 2
    • -f(x) + 2
    • 2f(x) - 3
  8. The maximum value of f(x) = q + p cos(x) graph is 7 and the minimum is 1. find p and q.

 

 

 

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