Could you PLEASE explain WHY a graph passes *through* the x-axis when the root/factor is of an odd multiplicity, and why a graph *bounces off* the x-axis when it's an even multiplicity?
WHY is the keyword. please please explain why!
and please no junk/useless answers either. please only answer if you know what you're talking about!
(sorry for using all these caps, but i'm a little desperate,,,)
2006-11-02 14:36:58 · 2 answers · asked by belleswan 3 in Science & Mathematics ➔ Mathematics
It has to do with the number of turning points a graph has and the end behavior. For example the graph of x^2-4x+4 "bounces off" the x-axis at (0,2). This graph's end behavior is it approaches + infinity on the right(the leading coefficient is +), and since it is of an even degree it must do the same on the left. If it went through the x-intercept, instead of bouncing off, it would approach - infinity on the left. The graph of x^3-6x^2+12x-8 goes through the x-intercept of (0,2). This graph's end behavior is it approaches + infinity on the right(the leading coefficient is +), and since it is of an odd degree it must do the opposite on the left. If it bounced on the x-intercept like that of x^2-4x+4, it would approach + infinity on the left instead of - infinity.
2006-11-02 14:51:30 · answer #1 · answered by futureastronaut1 3 · 0⤊ 0⤋
In short, (-1)^n = 1 when n is even, and (-1)^n = -1 when n is odd. If you take the nth real root of -1, it's 1 if n is even, and -1 if n is odd. Hence, a graph of even root "bounces off" the x axis, while one of odd root "goes through" the x-axis.
2006-11-02 22:43:22 · answer #2 · answered by Scythian1950 7 · 0⤊ 0⤋