How do you graph quadratics?

1 Answer
Feb 22, 2017

I think the most important thing is to understand how to complete the square. Once you have that down, graphing will be really easy.

Example: Graph #y = x^2 - 2x - 3#

We start by completing the square.

#y = 1(x^2 - 2x + m - m) - 3#

We need to convert the expression in parentheses into a perfect square. We do this using the formula #color(red)(m = (b/2)^2)# where #ax^2 + bx + c# is the expression in parentheses.

Calculating, we have that #m = 1#.

#y = 1(x^2 -2x + 1 - 1) - 3#

You're going to have to take out a negative #1#.

#y = 1(x^2 - 2x + 1) - 1(1) - 3#

#y = 1(x^2 - 2x + 1) - 4#

The expression in parentheses is a perfect square trinomial now.

#y = (x - 2)^2 - 4#

The most important point to know when graphing a quadratic is the vertex. In vertex form, #y = a(x - p)^2 + q#, the vertex is at #(p, q)#. Therefore, our vertex is at #(2, -4)#.

The next step is to set the initial function to #0# and finding the x-intercepts.

#0 = x^2 - 2x - 3#

#0 = (x- 3)(x + 1)#

#x = 3 and -1#

Now plot the two x-intercepts at #(3, 0)# and #(-1, 0)#. Although it will be implied by the placement of the two x-intercepts relative to the vertex, I'm going to talk about direction of opening a little bit.

In the form #y = a(x - p)^2 + q#:

•If #a > 0#, then #y# opens upwards (that's to say, the vertex is an absolute minimum)
•If #a < 0#, then #y# opens downwards (that's to say, the vertex is an absolute maximum).

You may want to create a table of values pertaining to the function. Select points (e.g. -3, -2, -1, 0, 1, 2, 3 ...) and evaluate them within the function.

You will at that point have enough information to create the graph.

Practice exercises

I'm sure your math teacher has already emphasized to you that practice is key to success in math. Here's some practice problems to make you more comfortable with everything shown in this answer.

#1#. Graph the following functions.

a) #f(x) = -x^2 - 4x + 5#

b) #g(x) = 1/2x^2 + 2x - 2#

Challenge problems

  1. Determine the value of #m# that will make the trinomial a perfect square.

#81/4x^2 + 18x + m#

  1. The roots of a quadratic function, of the form #ax^2 + bx + c = 0#, is given by the formula #x = (-b +- sqrt(b^2 - 4ac))/(2a)#. Prove this. Hint: you must complete the square and then solve for #x# . Note that this formula is often very important when finding the x-intercepts of a function, if the function is not factorable (e.g #x^2 + x - 1)#.

Hopefully this helps, and good luck!