Algebra Toolkit

1. We subtract 3 from the formula for \(f(x)\text{:}\) \(~~g(x)=\dfrac{1}{x}-3\)
2. We add 4 to \(x\) before applying the function. Think of replacing \(x\) by \(x+4\) wherever it appears in the formula for \(f(x)\text{:}\) \(~~g(x)=\dfrac{1}{(x+4)^2}\)
3. We multiply the formula for \(f(x)\) by \(\dfrac{1}{3}\text{:}\) \(~~g(x)=\dfrac{1}{3}\sqrt[3]{x}\)
4. We make the formula for \(f(x)\) negative: \(~~g(x)=-|x|\)

1. We replace \(x\) by \(x+1\) wherever it appears in the formula for \(f(x)\text{:}\) \(~~g(x)=(x+1)^2-3(x+1)\)
2. We multiply the formula for \(f(x)\) by \(-2\text{:}\) \(~~g(x)=-2x^3-2\sqrt{x}\)
3. We subtract 8 from the formula for \(f(x)\text{:}\) \(~~g(x)=-3-\dfrac{2}{x}\)
4. We replace \(x\) by \(x-2\) wherever it appears in the formula for \(f(x)\text{:}\) \(~~g(x) = \dfrac{4(x-2)-7}{(x-2)^2}\)

This transformation is a shift in the \(y\)-direction. We subtract 4 from each value of \(f(x)\text{.}\)

This transformation is a shift in the \(x\)-direction. It is helpful to insert another row into the table.

The basic function is \(~f(x)=\sqrt[3]{x}~\text{.}\) The output is multiplied by 5, and then 6 is added, so the transformations are

1. \(~g(x)=5f(x)~\text{,}\) so the basic graph is stretched vertically by a factor of 5
2. and \(~F(x)=g(x)+6~\text{,}\) so \(g(x)\) is shifted up by 6 units.

The basic function is \(~f(x)=\dfrac{1}{x^2}~\text{.}\) We subtract 1 from the input, then multiply the output by \(-3\text{,}\) so the transformations are

1. \(~g(x)=f(x-1)~\text{,}\) so the basic graph is shifted 1 unit to the right
2. and \(~G(x)=-3g(x)~\text{,}\) so \(g(x)\) is reflected about the \(x\)-axis and stretched vertically by a factor of 3.