Date: 01/16/97 at 23:59:49 From: Anonymous Subject: Graphing Parabolas I have trouble plotting points for a parabola. I find points from putting the original equation into standard form. I have trouble finding the vertex and the line of symmetry. I need a trick or a short-cut or something. Thank you.
Date: 01/18/97 at 23:09:58 From: Doctor Scott Subject: Re: Graphing Parabolas Hi! To find the vertex of a parabola, you probably learned a process called "completing the square" (shown below) which puts a quadratic function (a parabola) into standard form : y = a(x - b)^2 + c, where a, b, and c are constants. Once you have the equation in standard form, the vertex is at the point (b, c) and the line of symmetry is x = b. If a is positive, the parabola opens up, if a is negative, the parabola opens down. If the original equation is x = ay^2 + by + c, then it is a parabola which lies "on its side" and opens either to the left or the right. You can use completing the square to put the equation in standard form, x = c(y - h)^2 + k, with its vertex at (k, h) and line of symmetry y = h. To plot the parabola, I usually find that it is easiest to create a t-table (or a table of values) using a few points on either side of the vertex to find some points which lie on the parabola. Example: y = 4x^2 + 8x + 6 Complete the square: y - 6 = 4(x^2 + 2x ) <- move the constant and factor out a 4 Now find half the "x" term and square it, adding this to the right (inside the parentheses) and keeping the equation in balance by putting the appropriate amount on the left side. So, add 1 to the right, but put 4 on the left (because of the 4 outside the parentheses): y - 6 + 4 = 4(x^2 + 2x + 1) The right side is now a perfect square trinomial - so it factors to (x + 1)^2: y - 2 = 4(x - 1)^2 Finally, isolate y: y = 4(x + 1)^2 + 2 So, the vertex of the parabola is (-1, 2), the line of symmetry is x = -1 and to plot the parabola, create a small t-table: x | y ----------- -3 | -2 | -1 | 2 0 | 1 | Now, find the y-values for the table using the final form of the equation. (Note that we chose 2 points on the left of the vertex and 2 on the right). Hope this helps you out! -Doctor Scott, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum