Quadratic function

A quadratic function is a function that can be expressed by the prescription f(x) = ax² + bx + c, where a, b, c are the real numbers and further a ≠ 0. Just as a linear function is always described by a line, a quadratic function is in turn always described by a parabola.

An example of a quadratic function

An example of a simple quadratic function might be f(x) = x² + 3x − 7. The graph of this function would look like this:

A quadratic function could be written schematically as ax² + bx + c. The term ax² is called the quadratic term, and every quadratic function must have this term. The next term bx is called the linear term and it does not have to be in the quadratic function - it can be zero. For example, this function f(x) = 5x² + 7 is still a quadratic function. The last member of c is called the absolute member and is also optional. So for the function f(x) = x² + 3x − 7, the following would be true

$$ a = 1, b = 3, c = -7 $$

Why is a equal to one? Because the notation x² actually means 1 · x², hence a = 1.

Properties of the quadratic function

Starting with the definitional domain: the definitional domain of a quadratic function is ℝ, the set of real numbers, the range of values depends on the particular function, but it always goes to (plus or minus) infinity. Furthermore, the quadratic function is always increasing in the middle of the interval and decreasing in the other half. If the linear term is zero (b = 0), the quadratic function is even. A quadratic function is never a simple function.

Constraints from above or below

A quadratic function is always either bounded from above or from below. It depends only on the parameter a. This is because if the parameter a is positive, then the graph of the function "grows upwards", the graph looks like the letter "U" and the graph is thus bounded from below. An example is the function f(x) = 2x² with the graph:

We see that all function values (all red points) are above −1, so the function is bounded from below. If we had the function f(x) = −2x²:

then again all function values would be below the number 1, so the function would be upper bounded.

Convexity and concavity

Convexity and concavity again depend on the parameter a. A quadratic function is convex if it is in the shape of the letter "U" and is concave if it is in the shape of an inverted "U". So if a > 0, the graph is convex and if a < 0, the graph is concave.

The a parameter also affects whether the graph is narrow or wide. The closer the value is to zero, the wider the graph is and vice versa.

The intersections with the axis x and y

If we want to calculate the intersections of a function with the x or y axis , we just set up a simple equation and then solve it. We just need to remember that we can write the function in the form y = ax² + bx + c, which means that we can find the y-coordinate at the point x = 2 by adding two after all x.

For example, for the function y = x² − 4x + 3, we would get the y-coordinate 2² − 4 · 2 + 3 = −1 at the point x = 2. So the graph of the function certainly passes through the point [2, −1]. If we want to calculate the intersections with the axis x, then we are actually interested in the x-value at the point y = 0. The intersection with the axis x always has a y-coordinate of 0. Look at the following graph and it will become clear.

Therefore, in the notation y = x² − 4x + 3 we just substitute a zero after y and solve the equation. The resulting roots are the x-coordinates of the intersections.

$$ x^2-4x+3 = 0 $$

This is a quadratic equation that we can solve using standard procedures. For example, we can decompose it into the product form (x − 1) · (x − 3), from which we find that the roots are x₁ = 1 and x₂ = 3.

If we want to obtain the intercepts with the axis y, we follow the same procedure. We will take the notation y = x² − 4x + 3 and this time put a zero after x. Essentially, we are asking what is the functional value of the function at the point x = 0:

$$ 0^2-4\cdot0+3=3 $$

So the intersections with the x axis are [1, 0] and [3, 0], and with the y axis are [0, 3].

How to calculate the coordinates of a vertex

For a quadratic function, it is also very important to determine its vertex, and to avoid having to remember a rather complicated formula, we will show another way that uses the method of completing on a square. The vertex of a quadratic function is the point at which the function has a minimum or maximum. The function f(x) = x² − 4x + 3, whose graph looks like this:

has a vertex at the point [2, −1].

The standard notation for a quadratic function looks like this: f(x) = ax² + bx + c. We will convert this function to the form g(x) = (x + m)² + n, where [−m, n] is the vertex of the quadratic function. So for our function, we want to get the form (x − 2)² − 1. How do we get this shape?

First, we just write down the bracket: (x + m) and put half the value of the parameter b from the quadratic function after m. The parameter b is equal to minus four, so half is equal to minus two, so we get this notation: (x − 2)².

Now it's time for the second step, we need to subtract the excess entries. If you were to multiply this parenthesis, you wouldn't get the correct result, we wouldn't get the original function. We still have to subtract the square root of m from the parenthesis and add the parameter c from the original function.

So, specifically, subtract (−2)² from the parenthesis and add three (the c parameter ):

$$ (x - 2)^2 − 4 + 3 $$

After the adjustment, we get

$$ (x - 2)^2 − 1, $$

which is the final result. Since the function is of the form (x + m)² + n, where m = −2, n = −1, we know that the vertex has the coordinates [−m, n], so [2, −1].

We should still check if we have calculated correctly. We just converted the quadratic function x² − 4x + 3 to the form (x − 2)² − 1. But this new shape should describe the same function, so if we multiply the parenthesis, we should get the original shape of the function. So let's try multiplying (x − 2)² − 1.

$$\begin{eqnarray} (x - 2)^2 − 1 &=& (x-2) \cdot (x-2)-1\\ &=& x^2-4x+4-1\\ &=& x^2-4x+3 \end{eqnarray}$$

We can see that we got the same function, we made the adjustments correctly.