Basic Transformations on Graphs

Drawing the graph of y = f(x) + b, b ε R, from the known graph of y = f(x)

  

                    graph-of-y=f(x)

 

It is obvious that domain of f(x) and f(x) + b are the same. Let us take any point x0 in the domain of f(x). y|x=x0 = f(x0).

The corresponding point on f(x) + b would be f(x0) + b.

For b > 0 => f(x0) + b > f(x0) it means that the corresponding point on
f(x) + b would be lying at a distance 'b' units above point on f(x).

For b > 0 => f(x0) + b < f(x0) it means that the corresponding point on
f(x) + b would be lying at a distance 'b' units below point on f(x).

Accordingly the graph of f(x) + b can be obtained by translating the graph of f(x) either in the positive y-axis direction (if b > 0) or in the negative y-axis direction (if b < 0), through a distance |b| units.

Drawing the graph of y = |f(x)| from the known graph of y = f(x)

|f(x)| = f(x) if f(x) > 0 and |f(x)| = -f(x) if f(x) < 0. It means that the graph of f(x) and |f(x)| would coincide if f(x) > 0 and the portions where f(x) < 0 would get inverted in the upwards direction.

graph2

The above figure would make the procedure clear.

Drawing the graph of y = f(|x|) from the known graph of y = f(x)

It is clear that, f(|x|) = even-function

Thus f(|x|) would be a even function. Graphs of f(|x|) and f(x) would be identical in the first and the fourth quadrants (as x > 0) and as such the graph of f(|x|) would be symmetrical about the y-axis (as (|x|) is even).

 

                     graph3  

  

The figure would make the procedure clear.

Drawing the graph of |y| = f(x) from the known graph of y = f(x)

Clearly |y| > 0. If f(x) < 0, graph of |y| = f(x) would not exist. And if
f(x) > 0, |y| = f(x) would give y = + f(x). Hence graph of |y| = f(x) would exist only in the regions where f(x) is non-negative and will be reflected about the x-axis only in those regions.

               graph4

Drawing the graph of y = f(x + a), a ε R from the known graph of y = f(x)

           graph5

Let us take any point x0 ε domain of f(x), and set x + a = x0 or x = x0 - a.
a > 0 => x < x0, and a < 0 =>  x > x0. That mean x0 and x0 - a would given us same abscissa for f(x) and f(x + a) respectively.

As such, for a > 0, graph of f(x + a) can be obtained simply by translating the graph of f(x) in the negative x-direction through a distance 'a' units. If a < 0, graph of f(x + a) can be obtained by translating the graph of f(x) in the positive x-direction through a distance a units.

Drawing the graph of y = a f(x) from the known graph of y = f(x)

          graph6

  

It is clear that the corresponding points (points with same x co-ordinates) would have their ordinates in the ratio of 1 : a.

Drawing the graph of y = f(ax) from the known graph of y = f(x)

 

             graph7

 

Let us take any point x0 ε domain of f(x). Let ax = x0 or x = x0/a.

Clearly if 0 < a < 1 then x > x0 and f(x) will stretch by 1/a units against the y-axis, and if a > 1, x < x0, then f(x) will compress by a units against the y-axis.

Drawing the graph of y = f-1(x) from the known graph of y = f(x)

For drawing the graph of y = f-1(x) we have to first of all find the interval in which the function is bijective (invertible). Then take the reflection of y = f(x) (within the invertible region) about the line y = x. The reflected part would give us the graph of y = f-1(x).

e.g. let us draw the graph of y = sin-1 x. We know that y = f(x) = sin x is invertible if f : [-∏/2, ∏/2] → [-1, 1], => the inverse mapping would be f-1 : [-1, 1]→[-∏/2, ∏/2].

 

                graph8

 

Illustration:      Draw the graph of f(x) = cosx cos(x + 2) - cos2(x + 1).

Solution :        f(x) = cox cos(x + 2) - cos2(x + 1)

                      1/2 [cos(2x + 2) + cos 2]

 

                                        graph9 

 

                       -(1/2)[cos(2x + 2) + 1]

                     = (1/2) cos 2 -(1/2) < 0.

 

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