Posted by shansen1@pacbell.net (67.114.252.118) on October 03, 2003 at 21:51:08:
In Reply to: Graphing help! posted by on October 03, 2003 at 16:54:56:
Problem:
y = x^2 - 4
Solution:
The easiest way to solve this equation is by factoring.
Note that y = (x+2)(x-2).
The solution to this equation is where the graph crosses the x axis, i.e, where y = 0. This is called the 'x intercept'. As we travel along the graph (e.,g with our finger or pencil point) and cross the x axis, the y value for that crossing location must be zero. In this case, y equals zero (intercepts the x axis) whenever EITHER ONE of the above factors is zero. It doesn't matter if the other factor is not zero at the time. Zero times anything is still zero. So the solution to this equation (and the set of x intercepts) is x = {2,-2}.
Graph:
If you look in your book you will recall that the graph for the general 2nd degree equation
y = Ax^2 + Bx + C is in the shape of a parabola. The parabola opens either up or down depending on the sign of A. In this case A is positive (+2) so the parabola opens upward. This makes sense because as x becomes humongously positive or negative, the y value is positive (+ time + is +, etc.)
To define a parabola takes three points. So far we only have two, {2,0} and {-2,0}. Since the parabola opens upward and is symmetric about the y axis, a good point to establish would be the point where the parabola crosses the y axis, the 'y intercept.' To find this point we'll set x = 0, because we know that at that particular x value the graph is crossing the y axis SOMEWHERE, we just don't quite know where it is. To find out where, just plug x = zero into the equation.
y = x^2 -4
y = 0^2 -4 = -4
So a third point (the y intercept) would be located at {x,y} = {0,-4}
Although three points is all you need mathematically to completely determine a parabola, sometimes it helps to have a couple of more points, especially if you're a beginner at graphing. So lets obtain two more points by setting x = 4 and x = -4. I could have chosen a different value, like a hundred or something. But 4 is a small number that is easy to calculate, so I chose 4. A larger numbvber would make the graph go off scale too so that is why I chose a smalle number. Anyway, the y value for those two points (x= +/- 4) is y = 4^2 - 4 = 12. So this gives a total of five points which should be enough to draw a very nifty parabola that goes through all five points. To summarize, the points are {x,y} = {-4,12} {-2,0} {0,-4} {2,0} and {4,12}