Most people come out of an Algebra class capable of dealing with functions in the form \(y = f(x)\). However, many functions that you will have to deal with in a Calculus class are in the form \(x = f(y)\) and can only be easily worked with in that form. So, you need to get used to working with functions in this form.

The nice thing about these kinds of function is that if you can deal with functions in the form \(y = f(x)\) then you can deal with functions in the form \(x = f(y)\) even if you aren’t that familiar with them.

Let’s first consider the equation.

\[y = {x^2} - 6x + 5\]

This is a parabola that opens up and has a vertex of \(\left(3,-4\right)\), as we know from our work in the previous example.

For our function we have essentially the same equation except the \(x\) and \(y\)’s are switched around. In other words, we have a parabola in the form,

\[x = a{y^2} + by + c\]

This is the general form of this kind of parabola and this will be a parabola that opens left or right depending on the sign of \(a\). The \(y\)-coordinate of the vertex is given by \(y = - \frac{b}{{2a}}\) and we find the \(x\)-coordinate by plugging this into the equation. So, you can see that this is very similar to the type of parabola that you’re already used to dealing with.

Now, let’s get back to the example. Our function is a parabola that opens to the right (\(a\) is positive) and has a vertex at \(\left(-4,3\right)\). The vertex is to the left of the \(y\)-axis and opens to the right so we’ll need the \(y\)-intercepts (*i.e.* values of \(y\) for which we’ll have \(f\left( y \right) = 0\))). We find these just like we found -\(x\)-intercepts in the previous problem.

\[\begin{align*}{y^2} - 6y + 5 & = 0\\ \left( {y - 5} \right)\left( {y - 1} \right) & = 0\end{align*}\]

So, our parabola will have -\(y\)-intercepts at \(y = 1\) and \(y = 5\). Here’s a sketch of the graph.