PARABOLAS: STANDARD FORM, VERTEX FORM,
CARTESIAN FORM, OH MY!
A parabola is basically just a bowl shape, and it’s what a quadratic equation
looks like if you graph it. You’ll usually encounter quadratic functions in what’s
called standard form: f (x) = ax^2 + bx + c. Standard form is very user-friendly
because you can factor it or apply the quadratic formula to find the roots (we
covered this here). However, the new SAT will sometimes ask you to recognize a
different form, called vertex form: f (x) = a(x – h)^2 + k. Vertex form is useful for
graphing because the numbers h and k in the equation together give you (h,k)—
the coordinates of the vertex. The vertex is the minimum or maximum point (the
bottom or top of the bowl shape). You can easily tell if the vertex is a maximum
or a minimum by looking at the sign of your a value. If a is positive, the
parabola opens up, and the vertex is a minimum. If a is negative, then the
parabola opens down, and the vertex is a maximum. So it’s helpful to know how
to convert standard form into vertex form when you need to find that all-
important maximum or minimum point. Converting from standard to vertex
form involves a process called completing the square. Here’s how you do it.
Let’s start with a sample quadratic function in standard form: f (x) = x^2 + 4x – 5.
Let’s start by rewriting it with some space between the 4 and the –5:
f (x) = x^2 + 4x – 5What we want to do is make that x^2 + 4x part into a perfect square trinomial
by adding a number to it. We find that number by dividing the b constant (4 in
this case) by 2 and squaring the result. This gives us 2^2 = 4. We’re going to add a
4 and subtract a 4 in that space so that things balance out. (We can’t just add
something to an expression willy-nilly, since that would cause the algebra gods
to smite us. By subtracting the same number we added, we make things balance
out and keep the algebra gods happy.)
f (x) = x^2 + 4x + 4 – 4 – 5The first three terms form a perfect square trinomial, and it factors into two
identical quantities. Factoring just that part, we get f (x) = (x + 2) (x + 2) – 4 –5,
which we can rewrite as f (x) = (x + 2)^2 – 9. Voilà! We have completed the square
and converted from standard to vertex form. Remember that the constants h and
k will give us the vertex if we put them together. The 2 is positive, so to fit our
original vertex form, we have to switch the sign in order to get h = –2. So the x-
coordinate of the vertex is –2. The –9 is the k number, or the y-coordinate of the