Functions - Higher only – WJECStretches and compressions of graphs

If \(f(x) = x^2\), then \(af(x) = a(x^2)\). This tells us that we need to multiply each of the \(y\) coordinates on the graph by \(a\) in order to stretch the original graph.

\(f(x) = x^2\)

\(3f(x) = 3x^2\)

Looking at some coordinates, we know that \(f(x) = x^2\) goes through \(\left(-2, 4\right)\), \(\left(-1, 1\right)\), \(\left(0, 0\right)\), \(\left(1, 1\right)\), \(\left(2, 4\right)\) amongst other points.

For \(3f(x) = 3x^2\) we need to multiply each of the \(y\) coordinates by 3 to determine the stretch position of the graph.

\(\left(-2, 4\right) \rightarrow \left(-2, 12\right)\)

\(\left(-1, 1\right) \rightarrow \left(-1, 3\right)\)

\(\left(0, 0\right) \rightarrow \left(0, 0\right)\)

\(\left(1, 1\right) \rightarrow \left(1, 3\right)\)

\(\left(2, 4\right) \rightarrow \left(2, 12\right)\)

If \(f(x) = x^2\), then \(f(ax) = (ax)^2\). This tells us that we need to divide each of the \(x\) coordinates on the graph by \(a\) in order to compress the original graph.

\(f(x) = x^2\)

\(f(2x) = (2x)^2\)

Looking at the same coordinates again, for \(f(2x) = (2x)^2\) we need to divide each of the \(x\) coordinates by 2 to determine the compressed position of the graph.

\(\left(-2, 4\right) \rightarrow \left(-1, 4\right)\)

\(\left(-1, 1\right) \rightarrow \left(-0.5, 1\right)\)

\(\left(0, 0\right) \rightarrow \left(0, 0\right)\)

\(\left(1, 1\right) \rightarrow \left(0.5, 1\right)\)

\(\left(2, 4\right) \rightarrow \left(1, 4\right)\)