Theory

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

the quadratic story

We have investigated quadratic functions from several viewpoints. Time to collect all of our thoughts and characterize quadratic functions.

Quadratic Functions

A quadratic function is a function that can be represented by a formula of the form

\[ Q(t) = a \, t^2 + b \, t + c \]

where $a$, $b$, $c$ are real numbers and $a \ne 0$.

$\blacktriangleright $ Domain:

The natural domain of a quadratic function is all real numbers, $(-\infty , \infty )$. Of course, a particular quadratic function may be defined with a subset of the real numbers. We may refer to these as restricted quadratic functions.

Forms

Formulas for quadratic functions have three basic forms:

Standard Form: The standard form for a quadratic function looks like a sum of terms
\[ a \, t^2 + b \, t + c \]
Factored Form: The factored form for a quadratic function looks like a product of factors
\[ a \, (t - r_1)(t - r_2) \]
Vertex Form: The vertex form for a quadratic function looks like
\[ a \, (t - h)^2 + k \]

Each form serves a different purpose. For analysis purposes, the factored form reveals the zeros of the quadratic function and the vertex form reveals the extreme value of the function.

Note: So far, we have restricted our investigation to quadratics with real roots (or zeros). Therefore, we encounter quadratics that do not factor for us with real numbers. We call these irreducible quadratics. Later, in this course, we will bring in complex numbers and complete our characterization of quadratic functions. Complex numbers will allow us to factor all quadratics.

In the standard form $Q(t) = a \, t^2 + b \, t + c $,

$a \, t^2$ is called the leading term
$a$ is called the leading coefficient
$b \, t$ is called the linear term
$b$ is called the linear coefficient
$c$ is called the constant term

$\blacktriangleright $ Graph

The graph of a quadratic function is a parabola, which opens up or down. The direction depends on the sign of the leading coefficient.

$\blacktriangleright $ Zeros

Quadratic functions always have two roots. However, these roots can be complex numbers, which we will encounter later in this course. Restricting ourselves to only real numbers means that our quadratics have $0$, $1$, or $2$ real roots. These correspond to $0$, $1$, or $2$ intercepts on the graph, which correspond to $0$, $1$, or $2$ distinct factors over the real numbers.

Quadratics are a type of polynomial and zeros of polynomials are also called roots.

The two roots can be different (distinct) numbers corresponding to two different factors and intercepts on the graph.
The two roots can be the same number. In this case, the factorization is a square: $Q(t) = a(t-r)^2$. The root has an even multiplicity and the vertex of the parabola is the single intercept.
The two roots can both be complex numbers. There are no real roots. There are no intercepts on the graph. The quadratic does not factor over the real numbers.

$\blacktriangleright $ Continuity

All quadratic functions are continuous.

$\blacktriangleright $ End-Behavior

All quadratic functions have an even degree, $2$. That means they behave the same in their tails. They are unbounded. The sign of the leading coefficient will tell which way.

If the leading coefficient is positive, then

\[ \lim \limits _{t \to -\infty } Q(t) = \infty \]

\[ \lim \limits _{t \to \infty } Q(t) = \infty \]

If the leading coefficient is negative, then

\[ \lim \limits _{t \to -\infty } Q(t) = -\infty \]

\[ \lim \limits _{t \to \infty } Q(t) = -\infty \]

$\blacktriangleright $ Behavior

Quadratic functions exhibit one of two types of behavior. Which one depends on the sign of the leading coefficent.

If the leading coefficient is positive, then the quadratic function decreases and then increases
If the leading coefficient is negative, then the quadratic function increases and then decreases

Either way, the behavior switches at the critical number.

\[ Q(t) = a \, t^2 + b \, t + c \]

\[ Q'(t) = 2a \, t + b \]

$Q'(t) = 2a \, t + b = 0$ at $-\frac {b}{2a}$

If the leading coefficient is positive, then the quadratic function decreases on $\left ( -\infty , -\frac {b}{2a} \right )$ and increases on $\left ( -\frac {b}{2a}, \infty \right )$

If the leading coefficient is negative, then the quadratic function increases on $\left ( -\infty , -\frac {b}{2a} \right )$ and decreases on $\left ( -\frac {b}{2a}, \infty \right )$

$\blacktriangleright $ Global Maximum and Minimum

Quadratic functions have either a global maximum or a global minimum, not both.

If the leading coefficient is positive, then the quadratic function has a global minimum value of $Q\left ( -\frac {b}{2a} \right )$, which occurs at $-\frac {b}{2a}$.

If the leading coefficient is negative, then the quadratic function has a global maximum value of $Q\left ( -\frac {b}{2a} \right )$, which occurs at $-\frac {b}{2a}$.

$\blacktriangleright $ Local Maximum and Minimum

The only local extreme values are the global values.

$\blacktriangleright $ Range

If the leading coefficient is positive, then the range is $\left [ Q\left ( -\frac {b}{2a} \right ), \infty \right )$.

If the leading coefficient is negative, then the range is $\left (-\infty , Q\left ( -\frac {b}{2a} \right ) \right ]$.

and a nice graph...

We would like to see the parabola, the intercepts, the lowest/highest point, arrows, labelled axes, maybe an axis of symmetry.

$\blacktriangleright $ More about Zeros
Our quadratic functions have $0$, $1$, or $2$ real roots or zeros. These can be obtained from any of the three forms.

Standard Form: Given the standard form, $Q(t) = a \, t^2 + b \, t + c$, we can use the Quadratic Formula.
The zeros of $Q(t) = a \, t^2 + b \, t + c$ are

\[ \frac {-b + \sqrt {b^2 - 4 \, a \, c}}{2a} \, \text { and } \, \frac {-b - \sqrt {b^2 - 4 \, a \, c}}{2a} \]

$b^2 - 4 \, a \, c$ is known as the discriminant.
- If $b^2 - 4 \, a \, c > 0$, then there are two real zeros.
- If $b^2 - 4 \, a \, c = 0$, then there is one real zero, because the square root will be $0$.
- If $b^2 - 4 \, a \, c < 0$, then there are no real zeros. They will be complex numbers.
Factored Form:
Given the factored form, $Q(t) = a (t - r_1)(t - r_2)$, we can read off the roots or zeros as $r_1$ and $r_2$. Many times the leading coefficient, $a$, is not factored out. The zero product property allows us to set each factor equal to $0$ and solve.
Vertex Form:
Given the vertex form, $Q(t) = a (t - h)^2 + k$, we can set the formula equal to $0$ and solve.
\begin{align*} a (t - h)^2 + k & = 0 \\ a (t - h)^2 & = -k \\ (t - h)^2 & = -\frac {k}{a} \\ t - h & = \pm \sqrt {-\frac {k}{a}} \\ t & = \pm \sqrt {-\frac {k}{a}} + h \end{align*}
From $a (t - h)^2 + k = 0$, we can see that if $a$ and $k$ have the same sign, then there are no real solutions. The zeros are complex numbers.
Another perspective, is to factor a difference of two squares.
\begin{align*} a (t - h)^2 + k & = 0 \\ (\sqrt {a} (t - h))^2 + (\sqrt {k})^2 & = 0 \\ (\sqrt {a} (t - h) - \sqrt {k}) (\sqrt {a} (t - h) + \sqrt {k}) & = 0 \end{align*}
Either $(\sqrt {a} (t - h) - \sqrt {k}) = 0$ or $(\sqrt {a} (t - h) + \sqrt {k}) = 0$
Either $\sqrt {a} (t - h) = \sqrt {k}$ or $\sqrt {a} (t - h) = -\sqrt {k})$
Either $t - h = \sqrt {\frac {k}{a}}$ or $t - h = -\sqrt {\frac {k}{a}}$
Either $t = h + \sqrt {\frac {k}{a}}$ or $t = h - \sqrt {\frac {k}{a}}$

From the vertex form, we see that the zeros or roots include the term $\sqrt {-\frac {k}{a}}$. The inside of the square root is $-\frac {k}{a}$. This inside is negative if $a$ and $k$ have the same sign. This inside is positive if $a$ and $k$ have opposite sign.

There are real roots when $a$ and $k$ have opposite sign in vertex form.

Quadratic Analysis

Analyze $M(t) = -\frac {1}{2} (t-3)^2 + 5$

$M$ is a quadratic function, which means its graph is a parabola. The leading coefficient is negative, which means the parabola is opening down. The highest point on the parabola is $(3, 5)$. From this, we know there are two intercepts, which means two roots and two factors.

With these ideas, we can write an algebraic analysis.

Domain

The domain is $(-\infty , \infty )$, since $M$ is a quadratic function.

Zeros

The graph suggest that there are two real zeros (roots).

\[ M(t) = -\frac {1}{2} (t-3)^2 + 5 =0 \]

\[ M(t) = \frac {1}{2} (t-3)^2 = 5 \]

\[ M(t) = (t-3)^2 = 10 \]

\[ M(t) = (t-3)^2 - \left ( \sqrt {10} \right )^2 \]

a difference of two squares.

\[ M(t) = ((t-3) - \ \sqrt {10}) ((t-3) + \ \sqrt {10}) \]

\[ \text {Either } \, t - 3 - \sqrt {10} = 0 \, \text { or } \, t - 3 + \sqrt {10} = 0 \]

\[ \text {Either } \, t = 3 + \sqrt {10} \, \text { or } \, t = 3 - \sqrt {10} = 0 \]

$3 - \sqrt {10} \approx -0.1622776602$ and $3 + \sqrt {10} \approx 6.16227766$

These agree with the graph.

Continuity

$M$ is continuous, since it is a quadratic function.

$\blacktriangleright $ End-Behavior:

$M$ is a quadratice function with a negative leading coefficient. It is unbounded from below.

\[ \lim \limits _{t \to -\infty } Q(t) = -\infty \]

\[ \lim \limits _{t \to \infty } Q(t) = -\infty \]

$\blacktriangleright $ Behavior:

$M$ is a quadratice function with a negative leading coefficient. $M$ increases and then decreases, switching at the critical number, which is $3$.

$M$ increases on $(-\infty , 3]$. $M$ decreases on $[3, \infty )$.

$\blacktriangleright $ Global Maximum and Minimum:

$M$ is a quadratice function with a negative leading coefficient. $M$ increases and then decreases, switching at the critical number, which is $3$.

This tells us that there is a global maximum at $3$. The vertex form gives $5$ as the maximum.

End-Behavior lets us know there is no global minimum.

$\blacktriangleright $ Local Maximum and Minimum:

The global maximum is also a local maximums and that is the only local extrema that a quadratic can have.

$\blacktriangleright $ Range:

$M$ is a quadratic function with a maximum value of $5$. That makes its range $(-\infty , 5]$.

We can also get the other two forms for $M$.

$\blacktriangleright $ Factored form: $M(t) = -\frac {1}{2} (t - (3 - \sqrt {10})) (t - (3 + \sqrt {10}))$

$\blacktriangleright $ Standard form: $M(t) = -\frac {1}{2} t^2 + 3t + \frac {1}{2} = -\frac {1}{2}(t^2-6t-1)$

ooooo-=-=-=-ooOoo-=-=-=-ooooo
more examples can be found by following this link
More Examples of Quadratic Functions

2025-08-07 15:50:08