Quadratic function

Where a linear function is a first degree function, a quadratic function is a second degree function and is written in this form:

 

$$ \large f(x)=a \cdot x^2 + b \cdot x + c $$

 

\(\large a\) cannot be 0. If we insert 0 and multiply, it becomes nothing. The term disappears and what remains is a linear function

 

If you draw a quadratic function as a graph, it becomes a parabola, where the following applies:

 

  • If \(\large a>0\) the arms point upwards
  • If \(\large a<0\) the arms point downwards  
  • The parabola will always intersect the y-axis at \(\large (0,c) \)
  • \(\large a \) determines how steep the parabola is, the larger \(\large a \), the narrower the parabola becomes
  • The parabola is always symmetric about a vertical line through the vertex (the axis of symmetry)

 

parabola

 

Vertex and intersection

A quadratic function is related to quadratic equations.

To find the intersection with the x-axis, you must solve the quadratic equation where \( y=0\):

 

$$ \large 0=ax^2+bx+c $$

 

You must do as usual when solving quadratic equations.

The discriminant will tell you a little about the intersection of the parabola:

 

  • If \( D > 0\) it intersects the x-axis in two places
  • If \( D < 0\) it does not intersect the x-axis
  • If \( D = 0\) it intersects the x-axis only once

 

To calculate the vertex of a parabola, we also use the discriminant:

 

$$ \large D= b^2-4ac $$

 

When we have the discriminant, we can calculate \((\large x,y)\) for the vertex:

 

$$ \large x=\frac{-b}{2a} $$

$$ \large y=\frac{-D}{4a} $$

 

Example intersection

We try the function \(\large y=2x^2-2x+0\)

First, find the discriminant:

 

$$ D=b^2-4ac \Leftrightarrow $$

$$ D=2^2-4 \cdot 2 \cdot 0 \Leftrightarrow $$

$$ D=4 $$

 

The discriminant is positive.

This means that the parabola will intersect the x-axis in two places. We calculate the two places with the quadratic formula:

 

$$ \Large x= \frac{-b \pm \sqrt{d}}{2 \cdot a} $$

 

Intersection 1:

$$ \large x= \frac{-(-2) + \sqrt{4}}{2 \cdot 2} $$

$$ \large x= \frac{2+2}{4} $$

$$ \large \underline{\underline{x= 1}} $$

 

Intersection 2:

$$ \large x= \frac{-(-2) - \sqrt{4}}{2 \cdot 2} $$

$$ \large x= \frac{2-2}{4} $$

$$ \large \underline{\underline{x= 0}} $$

 

Example vertex

In the example we found that the discriminant is 4

Now we can calculate \((x,y)\) for the vertex:

 

$$ \large x=\frac{-b}{2a} \Leftrightarrow $$

$$  \large x=\frac{-(-2)}{2\cdot 2} \Leftrightarrow  $$

$$ \large x=0.5 $$

 

$$ \large y=\frac{-D}{4a} \Leftrightarrow $$

$$ \large y=\frac{-4}{4 \cdot 2} \Leftrightarrow $$

$$ \large y=-0.5 $$

 

Vertex = \((0.5, -0.5)\)

 

Coordinate sets

We now have three coordinate sets, the two intersections and the vertex:

 

$$ \large (0,0), (0.5, -0.5), (0,1) $$

 

We could draw the parabola with these three, but let’s find two more, by inserting -1 and 2 into the function:

 

$$ \large y=2 \cdot -1^2-2 \cdot -1+0 $$

$$ \large y=4 $$

 

$$ \large y=2 \cdot 2^2-2 \cdot 2+0 $$

$$ \large y=4 $$

 

\(x\) 0 0,5 0 -1 2
\(y\) 0 -0,5 1 4 4

 

Now it can be drawn.

parabola