4.5 Quadratic function

We call quadratic function a function $f$ for which there exist constants $a,b,c\in\mathbb{R}$, with $a \neq 0$ such that

\begin{equation}\label{eq-kvadraticka-fce}\tag{4.7} f(x) = ax^2 + bx + c \end{equation}

for every $x\in\mathbb{R}$. The domain of such a function is by definition the whole real line $\mathbb{R}$. The graph of a quadratic function is a parabola, see picture 4.4. The coordinates of the vertex of the parabola are easily revealed after squaring:

\begin{align} a x^2 + bx + c & \href{If you don't see why the two expressions are equal, try simplifying the expression on the right side of the equality. Do you get the expression on the left side?}{\class{mathpopup bg-info-subtle}{=}} a \Bigg( x^2 + 2 \cdot \frac{b}{2a} \cdot x + \Bigg(\frac{b}{2a} \Bigg)^{\!\!2}\, \Bigg) + c - \frac{b^2}{4a} = \nonumber \\ &= a \Bigg( x + \frac{b}{2a} \Bigg)^{\!\!2} + c - \frac{b^2}{4a}.\label{eq-ctverce}\tag{4.8} \end{align}

This adjustment is motivated by the simple requirement that the independent variable $x$ occurs only in a squared expression. This is accomplished with the clever addition and subtraction of quadratic terms as shown here.

The squared bracket in (4.8) is always non-negative. From there it follows that the vertex of the parabola is located at the coordinate point

\begin{equation*} \Bigg( -\frac{b}{2a}, \, c - \frac{b^2}{4a} \Bigg). \end{equation*}

From equation (4.8) it is evident that the sign of the coefficient $a$ decides whether all functional values are greater (smaller) than or equal to $c-\frac{b^2}{4a}$. The image set of the quadratic function is therefore

\begin{equation*} H_f = \begin{cases} \left[c-\frac{b^2}{4a}, +\infty \right), & a > 0, \\ \left( -\infty, c-\frac{b^2}{4a} \right], & a < 0. \end{cases} \end{equation*}

A well known formula applies to find the intersections $x_{\pm}$ of the function $f$ with the $x$ axis:

\begin{equation}\label{eq-pruseciky}\tag{4.9} x_\pm = \frac{1}{2a} \Big( -b \pm \sqrt{b^2 - 4ac} \Big). \end{equation}

The equation $ax^2 + bx + c = 0$ has therefore real solutions only under the assumption of non-negativity of the discriminant $b^2 - 4ac$.

The formula for the roots can be derived from a modification to a square. Looking for roots, i.e. solving the equation $ax^2 + bx + c = 0$ and by using the equality (4.8), we get

\begin{equation*} \left( x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}. \end{equation*}

From here, the solution can be expressed as follows:

\begin{equation*} x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2|a|}. \end{equation*}

Finally, by using the sign $\pm$, we can write in compact form

\begin{equation*} x_{\pm} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \end{equation*}

which is exactly (4.9).


At this point, it should be pointed out that there can be many different proofs of a claim. Some may be easier, some more complicated. For example, if we just wanted to validate the present statement, that is, $x_{pm}$ as given in (4.9) expresses the roots of a quadratic function (4.7), it is enough to proceed as follows40:

The validity of (4.9) can be easily verified by a simple substitution. Let us take $x_+$ and show that it is a root of (4.7).

\begin{equation*} \begin{aligned} a x_+^2 + b x_+ + c &= a \cdot \frac{1}{4a^2} \Bigg(\! -b + \sqrt{b^2 - 4ac} \Bigg)^{\!\! 2} + \frac{b}{2a} \Bigg( -b + \sqrt{b^2 - 4ac} \Bigg) + c = \\ &= \frac{1}{4a} \Bigg( b^2 - 2 b \sqrt{b^2 -4ac} + b^2 - 4ac \Bigg) - \frac{b^2}{2a} + \frac{b}{2a} \sqrt{b^2 - 4ac} + c = 0 \end{aligned} \end{equation*}

Thus $x_+$ is indeed a root! Analogously, it can be verified that $x_-$ is a root of (4.7), too.


Figure 4.4: Graphs of two quadratic functions.
Question 4.4

Let as take $a>b>0$. The numbers $a$ and $b$ are said to be in golden ratio41, if the ratio $\frac{a+b}{b}$ is the same as $\frac{a}{b}$. What is then the value of $\varphi =\frac{a}{b}$?

Show answer

$\displaystyle\varphi = \frac{1+\sqrt{5}}{2}$.