# Ask Uncle Colin: A Nonlinear Graph Transformation

Dear Uncle Colin,

I’ve figured out the graph of $y = e^x\br{2x^2 -5x + 2}$ using algebra and calculus; the second part of the question asks about $y = e^{x^2}\br{2x^4 - 5x^2 + 2}$. I feel like there must be a quicker way than redoing all of that work!

Some Heuristic Or Rule To Circumvent Unnecessary Toil?

Hi, SHORTCUT, and thanks for your message!

You can indeed circumvent unnecessary toil here – the trick is to notice that the right hand side of the second expression is the same as the first, but with all of the $x$s replaced by $x^2$s. Alternatively said, if the first equation is $y=f(x)$, then the second is $y = f\br{x^2}$.

We’re looking at graph transformations.

### Recap

Now, you’re likely familiar with linear graph transformations “inside the bracket”: given something like $y=f(2x)$, you need to “undo” the multiplication and halve the width of the curve. Given something like $y = f(x-3)$, you need to “undo” the subtraction and shift the curve three units to the right.

The reason for this is that when you want the $y$ value corresponding to a given $x$, you evaluate the function at the $x$-value after you’ve transformed it – so for the $y$ value of the curve where $x=-3$, you have to work out the value of the function with $x=-6$.

### What about the square?

Well, you need to undo it. Each of the points on the curve where $x>=0$ will correspond to two points on the new curve, where the new $x$ value is a square root of the old one.

So, your $x$-intercepts of $\frac{1}{2}$ and $2$ become $\pm \frac{1}{\sqrt{2}}$ and $\pm \sqrt{2}$, and your minimum at $\frac{3}{2}$ becomes $\pm \sqrt{\frac{3}{2}}$.

The $y$-intercept at $(0,2)$ stays put (and becomes a maximum).

Hope that helps!

- Uncle Colin

## A selection of other posts

subscribe via RSS