# Is it possible to check if transformed function is correct using test points?

Most of the (basic) math I've been doing in school has been easily checkable - it is immediately obvious if one had the right answer by just plugging it back in. Recently, doing function transformations, we were not taught a method to check if a transformed function was correct, and I am unable to figure it out myself.

Say we had an original function - $x^2-1%$ for example. I wanted to translate this function 2 units right, then reflect horizontally (over the y-axis). So I came up with the equation $(-x-2)^2-1$.

Plotting these two functions, with the original on the right:

http://i.stack.imgur.com/TKjiy.png (Image of the two graphs)

I can start at a point like $(0, -1)$, then add 2 units right, and reflect, and easily see that this transformation is correct. However, I would like to check algebrically if possible.

I've tried inputting the x coordinate in the transformed function and then trying to solve to see if I get the same y value, to no avail. I'm not sure how I can even guarantee a point exists on the original graph!

Summary: Given an original function, and a transformed function, is it possible to check if the transformed function satisfies applied transformations?

• @AndréNicolas - Oops, sorry, I meant reflect horizontally over the y-axis! Feb 28 '12 at 4:35
• To reflect across the $y$-axis, you replace $x$ by $-x$. So translation to the right of $y=x^2-1$ yields $y=(x-2)^2-1$ ($x=2$ behaves like $x=0$ used to). Then reflection in the $y$-axis gives $y=(-x-2)^2-1$, or equivalently $y=(x+2)^2-1$. Feb 28 '12 at 4:39
• @AndréNicolas - Ugh, another facepalm moment for me. Fixed. That was what I had graphed, and yet failed to write it properly. Thanks. Feb 28 '12 at 4:46
• The answer to the question in the title is yes, it is possible to use test points. They should not be too simple, and two are better than one. Of course you have to carry out the transformations on the test points correctly. You can also do a backwards check. If for example translation to the right by $2$ followed by reflection in the $y$-axis gets you a certain result, then reflecting in the $y$-axis and translating left by $2$ should get you back to the original. Feb 28 '12 at 4:51
• @AndréNicolas - Do you mind posting an example as an answer? I'm not sure how to use the test points - where do I substitute them into the functions? Feb 28 '12 at 5:03

In your particular example, your function is a polynomial of degree $n$ transformed to another polynomial. Your function $-x^2 - 1$ is a polynomial of degree 2 (also known as a quadratic). Whenever that is the case, you can rely on the unisolvence theorem, the fact that a polynomial of order $n$ is uniquely determined by $n + 1$ points. So if you test $n + 1$ points and they go where you expected them to go, you can be sure that you have performed the correct transformation.

• Thanks, knowing this was helpful! Feb 28 '12 at 5:37

The answer to the question in the title is yes, it is possible to use test points.

The test point(s) should not be too simple, and two are better than one. Of course you have to carry out the transformations on the test points correctly.

You can also do a backwards check. If for example translation to the right by $2$ followed by reflection in the $y$-axis gets you a certain result, then reflecting in the $y$-axis and translating left by $2$ should get you back to the original. Please note that going backwards involves undoing your transformations in the order opposite to the one that you used in the forward direction.

Here is an example with test points, using the curves that you discussed. So we start with $y=x^2-1$. We move the curve $2$ units to the right. So $2$ plays the role that $0$ used to, and we get the curve $y=(x-2)^2-1$. Then we reflect in the $y$-axis. Reflecting in the $y$-axis means that you replace $x$ everywhere by $-x$. Thus $y=(x-2)^2-1$ becomes $y=(-x-2)^2-1$. This many minus signs is kind of ugly, and can lead to error. So since $-x-2=-(x+2)$, we end up with the nicer equation $y=(x+2)^2-1$.

Now let's check, by taking a point on the curve $y=x^2-1$. Let $x=3$. Then $3^2-1=8$, so our test point is $(3,8)$.

Move this $2$ units to the right. We get $(5,8)$. Reflect this new point across the $y$-axis. We get $(-5,8)$. Now we ask, is $(-5,8)$ on the curve $y=(x+2)^2-1$? Substitute $-5$ for $x$ in $(x+2)^2-1$. We get $(-5+2)^2-1=9-1=8$. Check done!

• I think you should use the fewest number of simple points possible. EG, if you want to test that a degree 2 polynomial is equivalent to another, use 3 points. 0, 1, and -1 are great candidates. Feb 28 '12 at 5:29
• Thank you, this was immensely helpful. Having everything testable gives some security in knowing if I've gotten it right. @rrenaud Thank you for that tidbit of information. I presume it is the unisolvence theorem that minopret expanded on later. Thanks all! Feb 28 '12 at 5:41
• @CuriousFr: For a quadratic, it is true that $3$ test points will (if carried out correctly) give theoretically complete justification. This may not be of practical significance, since if you make an error you are likely to make the same kind of error again. Feb 28 '12 at 6:30