Vocabulary

- parabola
- quadratic function
- vertex form of a quadratic function

So far, we’ve mostly looked at linear functions, which in standard form were $ax+b$. Now we’ll shift to quadratic functions, which can be written as $ax^2+bx+c$, which when graphed form parabolas.

These are also known as degree two polynomials, since the highest exponent of a variable is two. By contrast, linear functions are degree one polynomials.

The parent function, the version with no transformations, of a quadratic is $f(x)=x^2$. Back in 1.2, we covered transformations and learned that you can manipulate functions to move, scale, and flip it. This leads us to the vertex form of a quadratic function.

\[f(x) = a(x-h)^2+k\]This should look familiar, and with any luck you can guess the effect of each of those variables. Both $h$ and $k$ determine the translations, and also where the vertex of the quadratic will end up. The vertex is the highest or lowest point of the parabola and also where it changes direction.

$a$ is your vertical scaling factor, and if negative, makes the parabola point downwards. Notice that we don’t include the horizontal scaling in this form. That’s because it would have an impact on the horizontal position of the vertex. Without it, you can identify the vertex by just recognizing $(h,k)$.

Let’s look at one and see what we can pick up without graphing it.

\[f(x) = 2(x-3)^2+4\]The $2$ tells us it’s been stretched vertically, but not flipped. The $3$ (watch your negative) and $4$ give us the vertex, which is at $(3,4)$. We should also look at the domain and range. Our domain, like any other quadratic, is all real numbers, but our range is $[4,\infty]$.

Working the other way, by being given key features, we can construct our function. Let’s make it a bit tricky by giving you just a vertex and another point, a $y$-intercept in this case: $(-2,3)$ and $-1$.

Let’s plug in what we can, which is $h$ and $k$, and then solve for $a$.

\[\begin{align*} y &= a(x+2)^2+3 \\ -1 &= a(0+2)^2+3 && \text{Substitute in other point.}\\ -1 &= 4a + 3 \\ a &= \frac{-1-3}{4} = -1 \end{align*}\]Our completed function is

\[f(x) = -(x+2)^2+3\]Vertex form has its uses, but standard form will be more desirable for what we’ll do with quadratics later. To convert, you’ll have to expand the square part of the function, which is multiplying two binomials.

\[(a+b)(c+d) = ac+ad+bc+bd\]Each term in one binomial is multiplied by every term in the other one. This is sometimes called FOIL for the pattern of first, outer, inner, last. That can be helpful, but don’t forget the underlying mechanic so you can expand the idea to trinomials later.

\[\begin{align*} f(x) &= (x+1)^2 + 3 \\ &= (x+1)(x+1) + 3 \\ &= x^2+1x+1x+1(1) + 3 \\ &= x^2 + 2x + 1 + 3 \\ &= x^2 + 2x + 4 \end{align*}\]I suggest squaring a number of those binomials yourself and see if you can pick up on a pattern.