Babydoll, these aren't radical expressions. They're quadratic equations. This might help you out some.
1).
y = x² - 5x + 3
We want it in the form y - k = a (x - h)², so we move the constant 3 to the left side of the equation and then complete the trinomial square on the right side of the equation. That enables us to write the right side as a binomial squared, which will result in the trinomial. This also helps us find h and k, the coordinates of the vertex of the parabola.
y - 3 + 25/4 = x² - 5x + (5/2)²
y + 13/4 = (x - 5/2)²
Vertex = (5/2, -13/4)
Table of Values - Read as (x, y).
(0, 3), (-1, 9), (-2, 17), (-3, 27), (1, -1), (2, -3), (5/2, -13/4), (3, -3), (4, -1), (5, 3)
2).
y = - x² + 6x -2
Get this equation in the same form as the other one. First, factor out the -1 in front. That means a = -1. So this parabola opens down.
y = - (x² - 6x + 2)
y + 2 = - (x² - 6x)
Now we complete the trinomial square on the right side of the equation. Since, in the process, we add -9 to the right side of the equation, we must also add -9 to the left side.
y + 2 - (-6/2)² = - [x² - 6x + (-6/2)²]
y + 2 - 9 = - (x² - 6x + 9)
y - 7 = - (x - 3)²
Vertex = (3, 7).
Table of Values - Read this the same way as the first one.
(0, -2), (-1, -9), (-2, -18), (-3, -29), (1, 3), (2, 6), (3, 7), (4, 6), (5, 3), (6, -2)
Unfortunately, I don’t have the capability to do graphics right now. You can find utilities on the Internet by simply googling for “graphing utility,” “graphing a parabola,” or something similar.
2007-07-05 17:17:03
·
answer #1
·
answered by MathBioMajor 7
·
0⤊
0⤋
You can download MathGV, a free graphing program, at the Web site here.
That will take care of the graph; it's a simple matter to plot some points. Just pick a value for x and solve for y. Since you don't say what a "suitable" table of values is, I don't know whether you need 20 per equation or 2 or 3 or what.
You can save the graphs as bitmap files and print them from Windows Paint or as .jpg files and print them from a photo viewing program.
2007-07-05 15:03:10
·
answer #2
·
answered by historian 4
·
0⤊
0⤋