Quadratic functions

Quadratic functions#

Note: this chapters is formed quickly from lecture notes. The figures are also missing.

Quadratic functions in the form of \(f(x)=ax^2+bx+c\), where \(a,b,c ∈ℝ\), are parabolas when poltted in (x,y)-plane.

For example

a) \(f(x)=3x^2-4x+1\) multiplies variable \(x\) squared by three, subtracts \(x\) multiplied by four and adds one.

The values can be computed depending on the value of variable \(x\), for instance

\(f(-1)=3\cdot(-1)^2-4\cdot(-1)+1=8\)

\(f(0)=3\cdot0^2-4\cdot0+1=1\)

\(f(1)=3\cdot1^2-4\cdot1+1=0\)

\(f(2)=3\cdot2^2-4\cdot2+1=5\)

  • if \(a < 0\), the parabola opens downwards, “a cap”

  • if \(a > 0\), the parabola opens upwards, “a cup”

  • if \(a = 0\), the functions becomes a linear function


Example. Solve \(f(x)=10\).

\(\begin{align}3x^2-4x+1=&10 \\ 3x^2-4x-9=&0 \end{align}\)

Using the quadratic formula

\(\begin{align}x=&\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=&\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot3\cdot(-9)}}{2\cdot3} \\ x=&\frac{4\pm\sqrt{124}}{6} \\ x=&\frac{4\pm\sqrt{124}}{6} \\ x=&\frac{4\pm2\sqrt{31}}{6} \\ x=&\frac{2\pm\sqrt{31}}{3} \\ \text{and so on...} \end{align}\)


How to find the roots?

When a function intersects the x-axis, the value of the function is 0. So the root can be found by solving an equation of \(f(x)=0\) by factoring or using the quadratic formula.

\(\begin{align}f(x)=&0 \\ 3x^2-4x+1=&0 \\ x=&\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ ...& \\ x=&\frac{1}{3} \ \text{or} \ x=1 \end{align}\)

The number of roots can be solved from the discriminant, \(D=b^2-4ac\)

  • if \(D > 0\), the function has two roots

  • if \(D = 0\), the function has one root and the root is at the same point as the vertex and root is \(x=-\frac{b}{2a}\)

  • if \(D < 0\), the function has no roots

The extrema: minimun and maximum#

The minimum and maximum value of the function can be found from the vertex or from the end points of a specific closed domain.

Let’s examine the function \(f(x)=3x^2-4x+1\).

Since \(a > 0\), the parabola opens upwards (cup). Therefore, the minimum value can be found from the vertex using formula of \(x=-\frac{b}{2a}\).

\(x=-{\frac{-4}{2\cdot3}}=\frac{2}{3}\)

Therefore the minimum is

\(\begin{align}f(\frac{2}{3})=&3\cdot(\frac{2}{3})^2-4\cdot\frac{2}{3}+1 \\ =&3\cdot\frac{4}{9}-\frac{8}{3}+1 \\ =&\frac{4}{3}-\frac{8}{3}+1 \\ =&-\frac{1}{3} \end{align}\)

Maximum value doesn’t exists, because \(a > 0\) UNLESS whe define the domain i.e. \(x ∈ [0,4]\) (or using inequality: \(0 ≤ x ≤ 4\)). In this case, the vertex is in the domain!

Let’s compute the value of our function in the points of the domain:

\(f(0)=3\cdot0^2-4\cdot0+1=1\)

\(f(4)=3\cdot4^2-4\cdot4+1=33\)

So the function has a maximum when \(x = 4\) and the maximum is 33.


Example of an open domain of \(]-1,1]\) (which is the same as \(-1 < x ≤ 1\)) .

Let \(gt(x)=-2x^2+6x-4{,}\ x\in]-1{,}1]\). Find the minimum and maximum.

The vertex is in

\(x=-\frac{b}{2a}=-\frac{6}{2\cdot(-2)}=\frac{3}{2}\), which doesn’t belong in the domain.

We can only find the value of the function in the end point of x = 1.

\(g(1)=-2\cdot1^2+6\cdot1-4=0\)

but we have to look at the point x = -1 as well.

\(g(-1)=-2\cdot(-1)^2+6\cdot(-1)-4=-12.\)

Concluding from the values of the function in the end points of the domain, the function has a maximum of 0 but no minimum.

The range is from minimum to maximum and in this case the range is \(]-12,0]\) (which is the same as -12 < x ≤ 0).


The set of all possible inputs of a function is called a domain. In other words, domain corresponds to the values of the variable x.

The set of all possible outputs of a function is called a range. In other words, range corresponds to the values that the function can have.

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