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 maximum value $c-{\frac {b^{2}}{4a}}$ is attained at point ${\frac {-b}{2a}}$. Case $a<0$: no local maximum 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 $x=-b/2a$ (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 that the function approaches infinity as $x\to \infty$ and as $x\to -infty$ Leading coefficient is negative (i.e., $a<0$) iff that 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 rule for differentiation of 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 rule for differentiation of 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:

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