We have seen that the trajectories of projectiles are described by quadratic equations. In fact, quadratic equations and formulas are used everywhere.

• 101 Uses of a Quadratic Equation Part 1
• 101 Uses of a Quadratic Equation Part 2

Quadratic Functions

Quadratic functions describe relationships between many different common measurements.

For us, heading to Calculus, quadratics are the simplest complex function. We want to understand rates of change and linear functions just don’t change their behavior. Quadratics change their behavior.

Being the simplest complex function makes quadratics our favorite for constructing compositions and illustrating the chain rule (Calculus).

We will see shortly, that involving quadratics in our investigation of rates of change will make their zeros of the utmost importance.

So, we would like to know everything about the zeros of quadratic functions.

To this end, we have three viewpoints into the inner workings of quadratic functions.

\(\blacktriangleright \) Standard Form Every quadratic function can be written as a sum, \(Q(t) = a \, t^2 + b \, t + c\), with \(a \ne 0\).

\(\blacktriangleright \) Vertex Form Every quadratic function can be written with one occurrence of the variable via a square, \(Q(t) = a \, (t - h)^2 + k\), with \(a \ne 0\).

\(\blacktriangleright \) Factored Form Every quadratic function can be written as a product, \(Q(t) = a \, (t - r_1)(t - r_2)\), where \(r_1\) and \(r_2\) are the real zeros and \(a \ne 0\) - provided the quadratic function has two real zeros.

When analyzing a quadratic function (or any function), we want to know two basic pieces of information: where is it and how is it changing?

• “Where is it?” includes the function’s zeros, maximum value, and minimum value.
• “How is it changing?” includes increasing and decreasing.

For a quadratic, all of this information is directly connected to its leading coefficient and zeros. So, we have many perspectives and viewpoints of a quadratic function’s zeros (also called roots).

Standard Form

The standard form of a quadratic function looks like \(Q(t) = a \, t^2 + b \, t + c\), with \(a \ne 0\).

From this form the zeros are given by the quadratic formula.

The quadratic formula tells us that this quadratic function has two roots. There are several possibilities.

\(\blacktriangleright \) Possible Root Types

The quadratic formula tells us that there are three possibilities for types of zeros or roots.

• two real roots
• one real root
• no real roots (two complex roots)

In each of the examples above the number of solutions was determined by \(\sqrt {b^2 - 4 a c}\). The inside of the square root, \(b^2 - 4 a c\), is called the discriminant and its sign tells us how many real solutions the equation has.

• If \(b^2 - 4 a c\) \(>\)\(=\)\(<\) \(0\), then there are two distinct real solutions.
• If \(b^2 - 4 a c\) \(>\)\(=\)\(<\) \(0\), then there is one real solutions.
• If \(b^2 - 4 a c\) \(>\)\(=\)\(<\) \(0\), then there are no distinct real solutions.

\(\blacktriangleright \) Where does the Quadratic Formula come from?

To answer this, we need a different form for a quadratic function, the vertex form.

Vertex Form

The vertext form of a quadratic function, \(f(x) = A \, (t-B)^2 + C\) presents some function information quickly.

For instance, the maximum or minimum value is the constant term, \(C\). This extreme valus of the function occurs in the domain at \(B\).

So, it can be advantageous, to convert from the standard form to the vertex form. This process is called completing the square.

Completing the Square (Algebraic Viewpoint)

Completing the square is an algebraic procedure. Later in this section, we will take a functional viewpoint of this (which will be more direct than the procedure below).

Zeros of linear functions were easy to identify. Simply apply a little algebra to the equation to get the variable by itself on one side of the equation. But, this is not always possible with equations of the form \(0 = a \, t^2 + b \, t + c \), because there are two occurrences of the variable and they have different degrees.

It will take a bit of Algebra to combine them into just one occurrence. This procedure is called completing the square.

We want

The only way that is going to happen is if \(A = a\). And, as long as we know \(A = a\), we can factor it out and work with what is left.

Then, let’s pretend our first step is factoring out \(A\) or \(a\) and let’s start over.

Step 1

Factor out the leading coefficent. In this way we can pretend that we had a monic quadratic from the beginning. That means we can pretend the leading coefficient was \(1\).

Starting over...starting with a leading coefficient of \(1\)...

Step 2

We want to complete the square on \(t^2 + b \, t + c\). When the leading coefficent is \(1\), then we call this quadratic monic.

\begin{align*} (t-B)^2 + C & = (t - B)(t - B) + C \\ & = t^2 - 2 \, t \, B + B^2 + C \end{align*}

This last line \(t^2 - 2 \, t \, B + B^2 + C\) should turn out to be \(t^2 + b \, t + c\).

We want

To complete the square with a monic quadratic, we need \(B = \frac {b}{-2} = -\frac {b}{2}\). Let’s put that in.

We want this to be \(t^2 + b \, t + c\).

leading terms match...check
linear terms match...check

Step 3

Now there is a mess for the constant term. We have \(\left ( -\frac {b}{2}\right )^2 + C\), when we just wanted \(c\).

Then let’s just pick \(C\) to be \(-\left ( -\frac {b}{2}\right )^2 + c = -\left (\frac {b}{2}\right )^2 + c\)

Whew!

Let’s see some examples

The whole point was to change from an expression with two occurences of the variable to an expression with only one occurence of the variable.

This makes solving for zeros much easier.

Instead of solving \(2 \, t^2 + 4 \, t + 6 = 0\), we can solve \(2 (t+1)^2 + 4 = 0\).

From this equation, we can see that this function has no zeros. Squares cannot equal negative numbers.

As we saw earlier, quadratic functions can have two real zeros, one real zero, or no real zeros.

The graph of \(y = 2 \, x^2 - 12\, x + 18\) would be a parabola touching the horizontal axis at only one point (its vertex).

There is another viewpoint on the example above. We arrived at this equation

which could be viewed as

If we proceed with the zero product property we would create two new equations. One for each factor.

Either \(x - 3 = 0\) or \(x - 3 = 0\).

Either \(x = 3\) or \(x = 3\).

\(3\) is the only solution to the equation, but the equation has this solution twice. A double root.

The previous two examples illustrated that quadratic equations can have two or one solutions. And, a quadratic equation can have no real solutions.

The above example illustrates that there must be numbers missing from the real numbers. We normally expect equations to have solutions.

\(\blacktriangleright \) A Peek Ahead

It feels like that quadratic equation should have a solution.

All we needed was for \((x-3)^2 = -1\)

There are no real numebrs that will do this. The square of a real number cannot be negative.

We can see that the real numbers are not enough for all of our equations. Eventually, we will fill in the missing pieces with the Complex Numbers. They will include \(\sqrt {-1}\). Then our last example will have two complex solutions.

• Our first example had two real solutions.
• Our second example had two identical solutions.
• This third example has two distinct complex solutions.

\(\blacktriangleright \) All quadratics will have two solutions...eventually.

We’ll fill in the holes later in this course.

For now, we are staying inside the real numbers.

For now, we note that there are no real solutions in the third example.

Completing the Square (Conceptual Viewpoint)

The previous viewpoint was an algebraic viewpoint.

It presented a step-by-step procedure for applying algebra that would transform a quadratic polynomial in standard form to one in verxtex form. The procedure gather terms that could be rewritten as a square.

The procedure began with the standard form, \(A \, x^2 + B \ x + C\), and the result of the procedure was an equivalent form, \(A (x - h)^2 + k\), called the vertex form.

The whole point was to change from an expression with two occurences of the variable to an expression with only one occurence of the variable.

This is helpful in graphing and helpful in solving quadratic equations.

However, the procedure isn’t the point of all of this. It is just a procedure to obtain the vertex form. If we take a functional viewpoint instead of an algebraic viewpoint, we can get there much faster.

Our goal is to convert \(a \, x^2 + b \, x + c\) into \(a \, (x - h)^2 + k\).

Let’s think of these as function.

• \(S(x) = a \, x^2 + b \, x + c\).
• \(V(x) = a \, (x - h)^2 + k\).

The \(a\) is the leading coefficient in both forms. Therefore, we only need \(h\) and \(k\) in terms of \(a\), \(b\), and \(c\).

\(\blacktriangleright \) The extrema

• From the standard form, we know that the vertex’s first coordinate is \(\frac {-b}{2 a}\).
• From the vertex form, we know that the vertex’s first coordinate is \(h\).

These must be equal.

If we evaluate each function at \(\frac {-b}{2 a}\), we must get the same value.

For \(\frac {-b}{2 a} = h\), we have

Therefore, if we evaluate \(S(x) = a \, x^2 + b \, x + c\) at \(\frac {-b}{2 a}\), we must also get \(k\).

This makes the example above much quicker.

These examples give us a procedure for developing the quadratic formula.

\(\blacktriangleright \) Where does the Quadratic Formula come from?

The discussion for the vertex form of a quadrtaic function tells us that the extreme value of the function happens at \(-\frac {b}{2a}\).

This means that to get the extreme value of the function, we need to evaluate at \(-\frac {b}{2a}\). We saw that this value is \(c - \frac {b^2}{4a}\)

All together the general form of the vertex for looks like

TO get the quadratic formula, we solve this equation for \(t\).

Either

or

• If \(a > 0\), then \(| 2a | = 2a\).
• If \(a < 0\), then \(| 2a | = -2a\).

Either way, we still get one negative and one positive fraction. Therefore, we can drop the absolute value signs.

Either

or

And, finally

People generally shorthand these two separate solutions as

This is known as The Quadratic Formula.

It gives the zeros or roots of a quadratic function where the standard form is

Now that we know what the zeros are of a quadratic function, we can factor it.

Factored Form

We have seen that there are at most two real solutions to \(a \, t^2 + b \, t + c = 0\), with \(a \ne 0\).

Let’s look at this from a function viewpoint. The quadratic function \(Q(t) = a \, t^2 + b \, t + c\) has at most two real zeros.

The quadratic formula gives explicit expressions for these roots.

\(\blacktriangleright \) Are there other quadratic equations, besides \(a \, t^2 + b \, t + c = 0\), that have these two solutions?

Perhaps, there are other quadratic functions, which have these two zeros. Let’s create a quadratic function from these two zeros.

If we multiply this out, we get

This quadratic function has the same two zeros as \(Q(t)\).

Note: If we multiply this by \(a\), we get

which is \(Q(t)\).

\(\blacktriangleright \) If two quadratics have the same two roots, then they must be multiples of each other.

We have discovered a new expression or formula or form for our quadratic functions.

The real zeros can be obtained via the quadratic formula - provided the discrimanant is positve. If the discriminant is \(0\), then \(r_1 = r_2\) and we get a square. If the discriminant is negative, then \(Q(t)\) doesn’t factor with real numbers.

The process of writing the function as a product is called factoring. \((t - r_1)\) and \((t - r_2)\) are the factors.

The quadratic formula always works. But, many times it is slow with lots of steps and reducing. Sometimes it is just easier to guess the factors. “Easier” usually means integers.

Guessing Factors

Let’s begin with a quadratic function written as a sum: \(g(x) = a \, x^2 + b \, x + c\), a.k.a standard form.

Step 1: If there are any common numeric factors among the three coefficients then factor them out.

Step 2: We are looking for a factorization \(g(x) = a \, x^2 + b \, x + c = (A \, x + B)(C \, x + D)\).

for this to happen, we need

• \(a = A \cdot C \)
• \(b = A\cdot D + B \cdot C\)
• \(c = B \cdot D\)

Therefore, look for pairs of factors of \(a\) and pairs of factors of \(c\).

Step 3: Step through your pairs of factors and look for \(b = A \cdot D + B \cdot C\).

As you gain more experience, this process becomes a mental process, which means it is fast.

Always give factoring a try, before moving onto the quadratic formula.

“Easier” usually means integers.

All of the factorizations above involved integers, because we are fast with those.

If the roots of the quadratic involve square roots, then we probably are not guessing them. In any case, the quadratic formula will find all of the roots.