Using Properties of Parabolas to Graph a Parabola

When graphing a parabola it is good to know that it is symmetric. A parabola will have a vertex. The vertex is the minimum or maximum value of the parabola. It is good to note, the range has a restriction on it, not the domain. The vertex represents the axis of symmetry as I discussed in a post about modeling data with quadratic functions.

Starting with the standard form of a parabola f(x)=ax^2+bx+c, you can determine the where the vertex will lie in the coordinate plane and other aspects of a parabola. Below are the properties of a parabola or a quadratic function in standard form.

Given that f(x)=ax^2+bx+c is a parabola when a\neq0.

  1. If a > 0, then the graph opens upward. If a < 0, then the graph opens downward.
  2. The axis of symmetry is given by the line -\frac{b}{2a}.
  3. The vertex is given by the ordered pair (-\frac{b}{2a},f(-\frac{b}{2a}))
  4. Finding 2 other points is easy. Choose any x-value and find the matching y-value. The second point is found by using the axis of symmetry and the location of the 1st points corresponding point.

The above information makes sketching the graph and finding the actual value of the maximum and minimum value of a parabola very easy. Just as you can write the equation of a parabola from three points, it is important to graph at least three points for a parabola.

Example Graphing a Quadratic Function

Graph -\frac{1}{2}x^2+2. To graph any quadratic it is important to get an idea of what it should look like from the properties. In this case a < 0. Which means the parabola, will open downward. Often, parabolas are referred to as u-shaped graphs, so the u will open downwards. The axis of symmetry (AOS) will be at

-\frac{b}{2a}=-\frac{0}{2*-\frac{1}{2}}=0.

This function does not have a linear term, therefore the AOS is given by x = 0. Any parabola without a linear term will have an AOS of x = 0. Finding the vertex requires you to substitute 0 in for the variable x. This is where function notation comes in handy,

f(0)=-\frac{1}{2}0^2+2

f(0)=2.

Thus, the vertex (-\frac{b}{2a},f(-\frac{b}{2a}))=(0,2)

Because parabolas are symmetric, you need to only find one point other than the vertex. The third point is found by finding the corresponding point of the second point found. In this case, you would want to find f(2), which means when x = 2. The related y-value would be found with the following steps.

f(2)=-\frac{1}{2}(2)^2+2  substitute 2 in for x

f(2)=-\frac{1}{2}(4)+2  Square the 2

f(2)=-2+2  Multiply

f(2)=0  Add

Figure 1 Parabola y = (-1/2)x^2+2

Figure 1 Parabola y = (-1/2)x^2+2

This gives a point of (2,0). The point the is symmetric about the AOS is (-2,0).All of the key points and the AOS are mark in Image 1 below. It is good to be worried about neatness when creating a graph, but if you are doing it by hand a good sketch will do. To the left is the best I could do using photo shop. If you need a an accurate graph, then using  graphing calculator or computer software would be your best bet.

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