Quadratic Function

The quadratic function is any function in the form

f(x) = ax2 + bx + c, where a ≠ 0.

Why the a ≠ 0 condition? Because if a = 0 then it’s actually a linear function not a quadratic one. The graph of a quadratic function is always a parabola. Let’s look at the graph of function f(x) = x2 + x + 1:

Function f(x) = x^2+x+1

Vertex

Each parabola has it’s vertex, which is a turning point. The vertex is either the minimum or the maximum of such parabola, depending on the constant a. See the following picture:

Function f(x)=x^2-4x+3 with marked vertex

The vertex is at the minimum of this function. It’s the point of the graph with the least y-coordinate. The question is how do we compute the coordinates of such point? We can use the technique of completing the square to rearrange the function equation to it’s alternate form. We get

f(x) = x2 − 4x + 3 = (x − 2)2 − 1.

From the last formula we can easily get the coordinates of the vertex. Let’s suppose the function f(x) = x2 for a moment. The graph follows:

Graph of f(x)=x^2

This function clearly has the vertex on the origin, i. e. the vertex has coordinates (0, 0). Now consider the function f(x) = (x − 2)2. How does it affect the graph? It moves the whole graph two points to the right:

Graph of f(x)=(x-2)^2

Since whole graph moved to the right, the vertex moved to the right too. Now the vertex has coordinates (2, 0). Let’s add the final part, the + 1 to the function definition: f(x) = (x − 2)2 − 1. We just move the graph one point downwards:

Graph of f(x)=(x-2)^2-1

The vertex moved downwards too so the vertex has coordinates (2, − 1). Because the function (x − 2)2 − 1 is the same as the original function x2 − 4x + 3 we can answer the original question: the function f(x) = x2 − 4x + 3 has vertex on coordinates (2, − 1).

To recapitulate. When finding the vertex of a quadratic function f:

  1. Convert the function f to the form y = a(x − h)2 + k using the completing the square technique.
  2. The vertex then has coordinates (h, k).

The Shapes of Parabola Based on Constants a, b and c

The constants a, b and c have different influence on the final shape of the parabola:

  1. The greater a the narrower the parabola is.

    Function f(x) = 2x^2 Function f(x) = 4x^2
  2. Parabola opens upward when a is positive and it opens downward when a is negative.

    Function f(x) = 2x^2 Function f(x) = -2x^2
  3. The positive c moves the graph upward, negative downward.

    Function f(x) = x^2+1 Function f(x) = x^2-1

Intersection Point With The Vertical Axis

The graph of each quadratic function f intersects the vertical axis. Why is that? Because a point on the vertical axis has coordinates in form (0, y). We can determine the y coordinate by computing the output f(0) of the quadratic function.

Example: let’s suppose a function f(x) = − 4x2 + 3x + 2.

Graph of f(x)=-4x^2+3x+2

Now we compute f(0). This is simple:

f(0) = 0 · x2 + 0 · x + 2 = 2.

The coordinates are (0, 2). We can even generalize it. Let’s suppose a quadratic function f(x) = ax2 + bx + c. Then f(0) equals

f(0) = 0 · x2 + 0 · x + c = c.

Thus for the quadratic function f(x) = ax2 + bx + c, the graph intersects the vertical axis in the point (0, c).

Intersection Point With The Horizontal Axis

The situation is different with the horizontal axis. The graph of a quadratic function can have zero, one or two intersection points with this axis. How do we find them? We simply solve the quadratic equation f(x) = 0. We get up to two results, x1 and x2. The coordinates of intersections points then are (x1, 0) and (x2, 0).

Basic Properties of Quadratic Function

  • The Domain of a quadratic function is always the whole set of real numbers.
  • The Image of a quadratic function depends on the coordinates of its vertex (h, k) and on the constant a.

    • if a > 0 then the image is the interval [k, ∞).

      Graph of function f(x) = 2(x-3)^2+1

      Because a > 0, the parabola opens upward and thus the lowest point of the parabola equals to the vertex. Thus all y-coordinates below this point are not part of the image of the function.

    • if a < 0 then the image is the interval ( − ∞, k].

      Graph of function f(x) = -2(x-3)^2+1

      The reasoning is the same but the parabola opens downward.