Quadratic function

Definition

A quadratic function is a function of the form:

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

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

Related notions

• Quadratic function of multiple variables
• L1-regularized quadratic function
• L1-regularized quadratic function of multiple variables

Key data

Item	Value
Default domain	all real numbers, i.e., all of $\mathbb {R}$
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 $C\in \mathbb {R}$.
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 $a,b,c\in \mathbb {R}$
$b^{2}-4ac$	(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 rots	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:

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

We get the answer:

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

Higher derivatives

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

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:

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

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.