**y = x ^{2} + 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

So, f(x) = x

Now whatever you put in as 'x' on the left, substitutes 'x' on the right.

- f(2) = 2
^{2}+ 3x2 -2 = 4 + 6 - 2 = 8 - f(-2) = (-2)
^{2}+ 3x(-2) -2 = 4 - 6 - 2 = -4 - f(0) = 0
^{2}+ 3x0 -2 = 0 + 0 - 2 = -2 - f(x+1) = (x+1)
^{2}+ 3(x+1) -2 = x^{2}+ 2x + 1 + 3x + 3 -2 = x^{2}+ 5x + 2 - f(x-1) = (x-1)
^{2}+ 3(x-1) -2 = x^{2}- 2x + 1 + 3x - 3 -2 = x^{2}+ x - 5 - f(2x) = (2x)
^{2}+ 3(2x) -2 = (2x)^{2}+ 6x -2 = 4x^{2}+ 6x - 2 - f(x/2) = (x/2)
^{2}+ 3(x/2) -2 = x^{2}/4 + 3x/2 -2 = x^{2}/4 + 3x/2 - 2

There are three major transformations:

**Translation****Reflection****Stretching**

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

f(x) -----> f(x+1) | f(x) -----> f(x-1) |
---|---|

translation in the -ve x-axis | translation in the +ve x-axis |

f(x) -----> f(x) + 2 | f(x) -----> f(x) - 2 |
---|---|

translation in the +ve y-axis | translation in the -ve y-axis |

You can animate the following curves to see how translation works. Please select an option and then move the sliders.

Choose slider, **a**, for the translation in the x-axis.

Choose slider, **c**, for the translation in the y-axis.

f(x) -----> f(-x) | f(x) -----> -f(x) |
---|---|

reflection in the y-axis | reflection in the x-axis |

f(x) -----> f(2x) | f(x) -----> 2f(x) |
---|---|

stretching parallel to x-axis | stretching parallel to y-axis |

You can animate the following curves to see how stretching works. Please select an option and then move the sliders.

Choose slider, **a**, for the stretching parallel to y-axis.

Choose slider, **c**, for the stretching parallel to x-axis.

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

**E.g.1**

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

- line of symmetry
- minimum value
- translations from f(x) = x
^{2} - solutions

First of all, let's use the completing the square method:

Let x^{2} + 2x -3 = (x+a)^{2} +b

x^{2} + 2x -3 = x^{2} + 2ax + a^{2}+ b

Make the coefficients of x^{2}, x and the constant on either side equal.

2a = 2 => a = 1

a^{2}+ 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) = x^{2}, 1 in the negative x-axis and 4 in the negative y-axis.

The solutions are, x = -3 and x = 1

**E.g.2**

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.

Transformation | New Coordinates | ||
---|---|---|---|

f(x+1) | A( -3 , 0 ) | B( -1 , 3 ) | C( 3 , -4 ) |

f(x-1) + 2 | A( -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) + 1 | A( 2 , 1 ) | B( 0 , 4 ) | C( -4 , -3 ) |

-f(x) + 2 | A( -2 , 2 ) | B( 0 , -1 ) | C( 4 , 6 ) |

Here is an opportunity for you to practise transformations using a sine curve. The curve is in the form of
`y = a + b sin(cx + d).` Change the values of *a, b, c and d* or to see the corresponding change. This is the best form of mastering the transformations.

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

- Use the completing the square method to sketch f(x) = x
^{2}+ 3x -4. Then describe the transformations of f(x) = x^{2}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) = x
^{4}- 2x^{2}+ 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.

- Show that the reflection in the y-axis of f(x) = x
- If f(x) = x
^{2}+ 2x, sketch the following:- f(-x)
- -f(x) + 2
- f(x+2) - 3
- f(2-x)
- 2f(x) + 1
- f(2x) - 3

- Sketch f(x) = x
^{2}-x - 6 and hence sketch g(x) = (x+2)(x+3) and show that f(x+1) = g(x). - 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?<
- Sketch f(x) = 1 + cos(3x) and hence find the period of the function, maximum value and minimum value.
- Sketch f(x) = 3 - 2sin(3x) and hence find the period of the function, maximum value and minimum value.
- Sketch f(x) = 1/x and then transform as follows:
- f(-x) + 2
- -f(x) + 2
- 2f(x) - 3

- The maximum value of f(x) = q + p cos(x) graph is 7 and the minimum is 1. find p and q.

Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7^{th} edition in print.