## Definition

A quadratic function is a function of the form:

$x \mapsto ax^2 + bx + c$

where $a,b,c$ are real numbers and $a \ne 0$. In other words, a quadratic function is a polynomial function of degree two.

Unless otherwise specified, we consider quadratic functions where the inputs, outputs, and coefficients are all real numbers.

## Key data

Item Value
Default domain all real numbers, i.e., all of $\R$
range Case $a > 0$: $\left[c - \frac{b^2}{4a},\infty\right)$
Case $a < 0$: $\left(-\infty,c - \frac{b^2}{4a}\right]$
period not a periodic function
local maximum value and points of attainment Case $a > 0$: No local maximum value
Case $a < 0$: local maximum value $c - \frac{b^2}{4a}$ is attained at point $\frac{-b}{2a}$.
local minimum value and points of attainment Case $a > 0$: local minimum value $c - \frac{b^2}{4a}$ is attained at point $\frac{-b}{2a}$.
Case $a < 0$: no local minimum value
points of inflection None
derivative The linear function $x \mapsto 2ax + b$
second derivative The constant function with constant value $2a$
$n^{th}$ derivative The first and second derivative are as described above. All higher derivatives are zero.
antiderivative $\frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + C$ with $C \in \R$.
important symmetry The graph of the function has mirror symmetry about the line $x = \frac{-b}{2a}$ (the vertical line through the unique critical point)
interval description based on increase/decrease and concave up/down Case $a < 0$:
increasing and concave down on $\left(-\infty,\frac{-b}{2a}\right]$
decreasing and concave down on $\left[\frac{-b}{2a},\infty\right)$
Case $a > 0$:
decreasing and concave up on $\left(-\infty,\frac{-b}{2a}\right]$
increasing and concave up on $\left[\frac{-b}{2a},\infty\right)$
power series and Taylor series The power series is the same as the polynomial, i.e., the power series about any point simplifies to the polynomial $ax^2 + bx + c$ (written in increasing order of powers of $x$ as $c + bx + ax^2$)

## Key invariants

Expression Name Significance in the case $a,b,c \in \R$
$b^2 - 4ac$ (unnormalized) discriminant The discriminant is positive (i.e., $b^2 - 4ac > 0$) iff the quadratic has two distinct real roots
The discriminant is zero (i.e., $b^2 - 4ac = 0$) iff the quadratic has a real root of multiplicity two
The discriminant is negative (i.e., $b^2 - 4ac < 0$) iff the quadratic has no real roots
$a$ leading coefficient Leading coefficient is positive (i.e., $a > 0$) iff the function approaches infinity as $x \to \infty$ and as $x \to -\infty$
Leading coefficient is negative (i.e., $a < 0$) iff the function approaches infinity as $x \to \infty$ and as $x \to -\infty$
$-b/a$ sum of roots If the roots are $\alpha,\beta$ (counted with multiplicity, and they need not be real roots), then the polynomial is $a(x - \alpha)(x - \beta)$ and the sum of roots $\alpha + \beta$ is $-b/a$.
$c/a$ product of roots If the roots are $\alpha,\beta$ (counted with multiplicity, and they need not be real roots), then the polynomial is $a(x - \alpha)(x - \beta)$ and the product of roots $\alpha\beta$ is $c/a$.
$b^2/(4a^2) - c/a$ normalized discriminant Similar observations as for the discriminant.

## Transformation

Any quadratic function can be expressed as a composite of a linear function, the square function, and another linear function. Explicitly, we can write:

$ax^2 + bx + c = a\left(x - \left(\frac{-b}{2a}\right)\right)^2 + \left(c - \frac{b^2}{4a}\right)$

In other words, it is a composite of three functions:

• $x \mapsto x - \left(\frac{-b}{2a}\right)$
• $x \mapsto x^2$
• $x \mapsto ax + \left(c - \frac{b^2}{4a}\right)$

## Differentiation

### First derivative

#### Computation as a linear combination of monomials

We can differentiate the polynomial termwise, using the fact that it is a linear combination of monomials:

$\frac{d}{dx}(ax^2 + bx + c) = a \frac{d}{dx}(x^2) + b \frac{d}{dx} x + c \frac{d}{dx} 1$

Now, using the differentiation rule for power functions, namely $d(x^n)/dx = nx^{n-1}$, we obtain:

$\frac{d}{dx}(ax^2 + bx + c) = a (2x) + b(1) = 2ax + b$

#### Computation using the transformed expression

We rewrote the polynomial as:

$ax^2 + bx + c = a\left(x - \left(\frac{-b}{2a}\right)\right)^2 + \left(c - \frac{b^2}{4a}\right)$

We differentiate:

$\frac{d}{dx}(ax^2 + bx + c) = a \frac{d}{dx}\left(x - \left(\frac{-b}{2a}\right)\right)^2 + \frac{d}{dx}\left(c - \frac{b^2}{4a}\right)$

The second expression, being constant, differentiates to zero. The first expression is a composite of $x \mapsto x - \frac{-b}{2a}$ and the square function. We can use the chain rule for differentiation to differentiate it. The answer we obtain is:

$\frac{d}{dx}(ax^2 + bx + c) = a \left[2\left(x - \frac{-b}{2a}\right)\right]$

Simplifying this, we get:

$\frac{d}{dx}(ax^2 + bx + c) = 2ax + b$

### Second derivative

#### Computation as a linear combination of monomials

We can differentiate the polynomial termwise, using the fact that it is a linear combination of monomials:

$\frac{d^2}{dx^2}(ax^2 + bx + c) = a \frac{d^2}{dx^2}(x^2) + b \frac{d^2}{dx^2} x + c \frac{d^2}{dx^2} 1$

Now, using the differentiation rule for power functions, namely $d^2(x^n)/dx^2 = n(n-1)x^{n-2}$, we obtain:

$\frac{d^2}{dx^2}(ax^2 + bx + c) = 2a + 0 + 0 = 2a$

#### Computation using the transformed expression

We rewrote the polynomial as:

$ax^2 + bx + c = a\left(x - \left(\frac{-b}{2a}\right)\right)^2 + \left(c - \frac{b^2}{4a}\right)$

We differentiate:

$\frac{d^2}{dx^2}(ax^2 + bx + c) = a \frac{d^2}{dx^2}\left(x - \left(\frac{-b}{2a}\right)\right)^2 + \frac{d^2}{dx^2}\left(c - \frac{b^2}{4a}\right)$

$\frac{d^2}{dx^2}(ax^2 + bx + c) = 2a$

### Higher derivatives

Since the second derivative is a constant (as we find using either of the computational methods above), all higher derivatives are zero.

## Points and intervals of interest

### Critical points

We can determine the critical point by setting the derivative to equal zero:

$2ax + b = 0$

This solves to $x = \frac{-b}{2a}$.

Alternatively, we can look at the form:

$f(x) = a\left(x - \frac{-b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)$

The critical point is therefore a shift by $-b/(2a)$ of the critical point for the square function, which is known to be at $0$.

### Points of inflection

The second derivative is the constant function with value $2a$, which is never zero. Therefore, there are no points of inflection.

### Local and absolute extreme values

#### Case a > 0

In this case, the unique critical point $x = -b/(2a)$ defines a point of local minimum, and it is also the unique point of absolute minimum. This can be seen in any of the following equivalent ways:

Method Description Comment on deducing local versus absolute minimum
transformed description of function We have $f(x) = a\left(x - \frac{-b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)$. The constant term $\left(c - \frac{b^2}{4a}\right)$ does not affect the point of local minimum. The expression $a\left(x - \frac{-b}{2a}\right)^2$ is always nonnegative (because $a > 0$). It is zero iff $x - \frac{-b}{2a} = 0$, or $x = \frac{-b}{2a}$. This, therefore, is the unique point of local and absolute minimum. Note that this method directly gives us that it is the unique point of absolute minimum.
first derivative test The derivative function $x \mapsto 2ax + b = 2a\left(x - \frac{-b}{2a}\right)$ is negative for $x < \frac{-b}{2a}$ and positive for $x > \frac{-b}{2a}$. Therefore, $f$ is decreasing on the left and increasing on the right of $-b/(2a)$, and thus has its unique local and absolute minimum at this point. The first derivative test, as conventionally stated, only talks about the sign of the derivative on the immediate left and right, and in this form, suffices only to deduce local extreme behavior. However, in this case, we have global information on the sign of derivative on the left and right of the point, therefore we can deduce that we have an absolute (global) minimum.
second derivative test The second derivative is the positive number $2a$ everywhere, so the critical point is a point of local minimum, and also the unique point of absolute minimum. The second derivative being positive at the critical point suffices to deduce that it is a point of local minimum. The second derivative being positive everywhere tells us that the function is a convex function (i.e., the graph is concave up), and therefore, any local minimum is unique and equals the absolute minimum.

#### Case a < 0

In this case, the unique critical point $x = -b/(2a)$ defines a point of local minimum, and it is also the unique point of absolute minimum. This can be seen in three ways, analogous to the case $a > 0$.

## Integration

### First antiderivative

We can use that indefinite integration is linear to convert this to a problem of integrating power functions:

$\int (ax^2 + bx + c) \, dx = a \int x^2 \, dx + b \int x^1 \, dx + c \int x^0 \, dx$

We can now use the integration rule for power functions ($\int x^r \, dx = \frac{x^{r+1}}{r + 1} + C$) and obtain:

$\int (ax^2 + bx + c) \, dx = a \frac{x^3}{3} + b \frac{x^2}{2} + cx + C, C \in \R$

### Higher antiderivatives

The process of integration can be repeated. The following is the formula for the $k$-fold integral of the function:

$a \frac{x^{k+2}2!}{(k + 2)!} + b \frac{x^{k+1}}{(k + 1)!} + c \frac{x^k}{k!} + \mbox{arbitrary polynomial of degree at most } k - 1$

The arbitrary polynomial of degree at most $k - 1$ is parameterized by $k$ coefficients. These are the constants of integration that accumulate, one in each step of integration.