Warm-up solution

Here is a graph of a function, \(f(x)\).

Cubic shaped graph crossing the x-axis at -1 and touching it at 2, and crossing the y-axis at 4

For each of the following transformations, sketch the transformed graph and write its equation in terms of \(f\).

Translation by \(\begin{pmatrix}-2\\0\end{pmatrix}\)

This is a translation to the left by \(2\) units.

The translated graph crosses the x-axis at -3, has its local maximum at (-2,4) and its local minimum at the origin

Algebraically, when translating \(2\) units left we replace \(x\) with \(x+2\), so we can write \[y=f(x+2).\]

Translation by \(\begin{pmatrix}0\\-4\end{pmatrix}\)

The graph is translated \(4\) units downwards.

The translated graph has its local maximum at the origin and its local minimum at (2,-4)
We can write this as \[y=f(x)-4.\]

You can think of this as replacing \(y\) with \(y+4\), which is consistent with how we treated \(x\) in the first example.

Stretch by factor \(3\) parallel to \(x\)

The graph is made \(3\) times as wide.

The stretched graph still has its maximum at (0,4) and its minimum is now at (6,0)

We replace the \(x\) with \(\frac{x}{3}\), so we can write this as \[y=f\left(\frac{x}{3}\right).\]

Stretch by factor \(\frac{1}{2}\) parallel to \(y\)

The graph is made half as tall.

The stretched graph still has its minimum where it touches the x-axis at 2 but now the maximum is at (0,2)

We replace \(y\) with \(2y\) and write this as \[y=\frac{1}{2}\;f(x).\]