Suppose that you have the graph of \(y = f\left( x \right)\). How can you draw the graph of \(y = g\left( x \right) = f\left( {2 x}\right)\)? Suppose that \(\left({{x_0},{y_0}} \right)\)is a point on the curve of the original function.This means that \(f\left( {{x_0}} \right) ={y_0}\). For what value of *x* will *g* output this same value \({y_0}\)? Clearly, it will be \(\frac {{{x_0}}}{2}\), as the following shows:

\[g\left( {\frac{{{x_0}}}{2}} \right) = f\left({2\left( {\frac{{{x_0}}}{2}} \right)} \right) = f\left( {{x_0}} \right) ={y_0}\]

This means that a point \(\left( {{x_0},{y_0}} \right)\) on the curve of *f* will correspond to the point \(\left({\frac{{{x_0}}}{2},{y_0}} \right)\) on the curve of *g*. Thus, every point on the original graph will come closer to the *y*-axis by a factor of 2 on the new graph.

In the following figure, the dotted curve corresponds to \(y = f\left( x \right)\)(where *f* is some function), while the solid curve corresponds to \(y = g\left( x \right) = f\left( {2x} \right)\):

Note how the curve has scaled(compressed) horizontally by a factor of 2. For example, the peak at \(\left( {1,3} \right)\) has shifted to \(\left( {0.5,3} \right)\) in the new curve.

The following figure shows the graph of \(y = f\left( x \right)\) and \(y = f\left( {3x} \right)\) (where *f* is some function):

Note how the graph of \(y = f\left( {3x} \right)\) is thrice as fast the graph of \(y = f\left( x \right)\). In once *cycle* of the original curve, the new curve completes three cycles(highlighted by the shading). In other words, the graph of \(y = f\left( {3x} \right)\) can be obtained by horizontally compressing the graph of \(y =f\left( x \right)\) by a factor of 3.

The following figure shows the graph of \(y = f\left( x \right)\) and \(y = f\left( {\frac{x}{2}} \right)\) (where *f *is some function):

Clearly, the graph of \(y = f\left( {\frac{x}{2}} \right)\) has scaled horizontally by a factor of 2 relative to the graph of \(y = f\left( x \right)\).

In general, the graph of \(y = f\left( {kx} \right)\) can be obtained by *horizontally scaling* the graph of \(y = f\left( x\right)\) by \(\left| k \right|\)units. If the magnitude of *k* is greater than 1, then the graph will *compress horizontally; *if the magnitude of *k* is less than 1, then the graph will *stretch out horizontally*. What happens if *k* is negative? The following figure shows the graph of \(y =f\left( x \right)\) and \(y = f\left( { -x} \right)\):

The two are mirror images of each other in the *y*-axis. In other words, the originally curve can be reflected about the *y*-axis to obtain the new curve. Can you see why?

The following figure shows the graph of \(y = f\left( x \right)\) and \(y = f\left( { - \frac{x}{2}} \right)\):

This transformation can be thought of as a combination of reflection about the *y*-axis and then horizontal scaling:

Note that in horizontal scaling, the point where the new curve intersects the *y*-axis is the same as the point where the old curve intersects the *y*-axis. To see why, suppose that the curve for \(y = f\left( x \right)\)crosses the *y*-axis at \(\left( {0,{y_0}}\right)\). This means that \(f\left( 0\right) = {y_0}\). If the horizontally scaled curve corresponds to \(y = g\left( x \right) = f\left( {kx} \right)\),then

\[g\left( 0 \right) = f\left( {k\left( 0 \right)}\right) = f\left( 0 \right) = {y_0}\]

Thus, the curve for *g* also passes through \(\left( {0,{y_0}} \right)\).