Vocabulary

- standard form of a quadratic function

Standard form, which looks like $ax^2 + bx + c$, is the version you will see most often, so we’ll spend some time learning how to work with it.

We went from vertex to standard form last section, and it wasn’t too bad. But going the other way is trickier. So let’s instead vertex form to help us out.

Here’s what vertex form looks like when made into standard form.

\[\begin{align*} f(x) &= a(x-h)^2 + k \\ &= a(x^2-2hx+h^2) + k \\ &= ax^2 - 2ahx + ah^2 + k \\ \end{align*}\]OK, bit of a mess of variables, but it’s all there. Given any standard form quadratic we can pick up our $a$, $h$, and $k$ without too much trouble.

$a$ is easiest. Look at your $x^2$ term and $a$ is whatever coefficient is there. And it’s also no coincidence that $a$ is used in both vertex and standard form.

Our $x$ term is used to find $h$. Its coefficient $b$ is defined as $-2ah$, so

\[\begin{align*} b &= -2ah \\ -\frac{b}{2a}&=h \end{align*}\]The horizontal component of our vertex we can find by taking $b$ and dividing it by $2a$ and remembering to flip the sign.

Now, we could look at $c$, which is defined as $ah^2 + k$, in order to find $k$, or we could just plug our newly found horizontal component into the function to find the vertical. One less equation to remember.

Let’s try one out. $x^2 - 6x + 10$.

$a$ is 1. Easy start. $h$ is $\frac{-b}{2a}$, so $\frac{6}{2}$ or 3 in this case. And to find $k$, we plug into the original. $f(3) = (3)^2 - 6(3) + 10 = 1$. Our vertex is $(3,1)$, giving us a vertex form of

\[f(x)= (x-3)^2+1\]As an added bonus, our $y$-intercept is just our constant term $c$, similar to linear functions.

With a line, you need two points to define it. With a quadratic, you need three. Given any three points, you can construct a system of equations to find the quadratic. The goal is to find $a$, $b$, and $c$, so by plugging in your three $x$ and $y$ coordinates, you get three different equations.

I won’t retype it here, but the example in the book shows how this is done. Use your matrix tools when its ready, and you’ll have your coefficients.

Regression is used to make predictions based on data. It takes the information that already exists and tries to make a line, or curve, of best fit. The situation determines what model is more appropriate. In the case of projectile motion, quadratic regression is ideal.

I will demo how to do this in class in Desmos, but the quick steps are add a table and then add regressions using the button in the top-left. Here is the book’s example recreated in Desmos, and also the help article on using regression in Desmos.

As for the $R^2$ value that will be displayed, that just tells you how well the existing data fits the model. The closer to 1, the better (which is a gross oversimplification).