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Graphing Quadratic Functions
Recall that any function that can be written in the form of f (x) = ax^2 + bx + c , where a, b, and c are real numbers and a ≠ 0 are quadratic functions. The graph of a quadratic function yields the shape of a parabola. If the value of the “ a” term is positive the parabola will open upward, whereas if the value is negative it opens downward.
The lowest point of a graph opening upward (the minimum ), or the highest point of a graph opening downward (the maximum ) is known as the vertex. An axis of symmetry can be determined by drawing a vertical line through the vertex of a parabola. This means that the equation for the axis of symmetry will be equal to the x value of the vertex. For example, if the vertex of a parabola was ( 1 , 3 ) , the formula for the axis of symmetry would be x = 1. This is without regard to the direction, up or down, that the parabola opens.
Opens upward Opens downward
As seen in earlier sections, the process of completing the square is a useful tool in finding non- integer values of quadratic equations, especially intercepts. When a quadratic equation of the form f (x) = ax^2 + bx + c is put through the process of completing the square it yields an equation of the form f (x) = a(x – h)^2 + k. The conversion of the equation to this form will yield critical information about the equation’s characteristics before you begin to graph it.
1.) The value of h is the distance left (if negative) or right (if positive) the graph translates from the standard position. 2.) The value of k is the distance up (if positive) or down (if negative) the graph translates from the standard position. 3.) The values of h and k , when put together as an ordered pair, give the vertex i.e. (h, k). 4.) The equation x = h is the formula for the axis of symmetry.
The following example demonstrates how to find the following critical information of the equation:
a.) vertex b.) axis of symmetry c.) y intercept (if any) d.) x intercepts (if any)
Example 1: Find the vertex, axis of symmetry, x-intercept(s), and y-intercept and graph the equation f(x) = x^2 + 2 x - 1.
Solution:
Step 1: y-intercept
In the form f (x) = ax^2 + bx + c , (0, c ) is the y intercept. In this example, for this step, we need to rewrite the given function to the proper form to get this information.
f (x) = x^2 + 2 x – 1 f (x) = x^2 + 2 x + ( - 1)
Therefore the y-intercept is: (0, -1)
Example 1 (Continued):
Step 5: Find the “reflection” of the y intercept and graph.
All of the points of a parabola have points that are reflections of each other across the AOS (Axis of Symmetry). Notice that the x-intercepts are a distance of from the x value of –1, the AOS. To find the reflection of the y intercept, duplicate the y value of the point and find the x distance to the AOS then travel the same distance on the other side of the AOS. In this case, the y value of the reflection of the y intercept, (0, -1) is –1, so the reflected point will also have a y value of –1. From the x value of the y intercept, 0 , it is a distance of 1 to the AOS value of –1. The value that is the same distance on the other side of the AOS is –2. Therefore, the reflected point of the y intercept is (-2, -1). Using all of this information, plot your points and graph.
Step 6: Graph.
