Basic Transformations on Graphs

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

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|) = 

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).

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.