Vertex Form Of Quadratic Functions: Fill & Download for Free

This is a common problem that troubles most people at first.

First I’ll type out the formula. It’s worth doing at least once. If [math]a[/math], [math]b[/math], and [math]c[/math] are real numbers then:[math]ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]One reason that it’s significant is because of the simplicity. It’s hard to think of it as simple when you’re first learning and working with it, but think about what it gives us: no matter which quadratic function you give me, I can always find where it crosses the [math]x[/math]-axis. ALWAYS. This is an extremely hard problem for other types of functions.Unless [math]x[/math] is a very particular number, then [math]\sin(x)=0[/math] is probably impossible to express using radicals and other simple operations. Even for cubic polynomials there is a so-called cubic formula, which is a formula that finds the root of a cubic function. But it’s VASTLY more complex than the quadratic formula; just writing it on a whiteboard would probably take a whole board. The quartic formula is, again, enormously complex compared to the cubic and quadratic versions. And for a quintic, AKA degree [math]5[/math], function, it is impossible to find such a formula using roots and other algebraic operations. It’s not just that we haven’t been clever enough to find one - it was proven that such a formula which works for every quintic is impossible (called the Abel-Ruffini theorem). It’s incredible that a simple polynomial can be unsolvable like that.So, one reason that the quadratic formula is significant is its existence and simplicity.Another significant aspect of the formula is the form of it. For one thing, we know of quadratics that don’t cross the [math]x[/math]-axis; [math]x^2+x+2[/math] is such a quadratic. That means that no real values of [math]x[/math] satisfy [math]x^2+x+2=0[/math]. The quadratic formula in fact predicts this: using the formula we encounter the number [math]\sqrt{-7}[/math]. What real number squares to [math]-7[/math]? Of course the answer is “none”. This is true for any negative number.For this reason the quantity [math]b^2-4ac[/math] is very important when studying quadratic functions. This is called the discriminant [math]\Delta[/math] (a capital D in Greek) because it discriminates (tells the difference between) different types of quadratics. Namely, if [math]\Delta<0[/math], then the square root in the formula is not a real number, meaning that the quadratic has no real number roots (although it does have complex roots).The discriminant tells us even more. If [math]\Delta=0[/math], then where the formula would normally have two answers due to the [math]\pm[/math], we actually only have one. This is called a double root, and it’s immediately clear from the formula when a quadratic has a double root. Of course, if [math]\Delta>0[/math] then we are in the “nice” case where the quadratic has two distinct real roots.Right away, the formula has allowed us to break ALL quadratics into three important categories. That’s significant to me.Another minor observation - let’s break the formula into two parts:[math]x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}[/math]The first part might be familiar: this is the [math]x[/math]-coordinate for the vertex of the graph of the quadratic. Then the formula tells us that the two roots are equally spaced from the vertex - one lower, and one greater. This is an amazing insight that you might not have ever had without the formula.So, why is it significant? It gives us insight into the three types of quadratic functions and tells us something fascinating about their shape.

What is the vertex of the quadratic function y=(x+7)²-8?Vertex form of a quadratic is a(x - h)² + k. In vertex form the vertex is (h,k)So the vertex is (-7,-8)