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Or another way is to memorize that the center and bottom of an upward facing parabola is at location x=4 because it is one half and negative of the minus eight x. Then substitute 4 for x and determine how far below the x-axis the bottom of the parabola is. Hence y=16-32+5 or negative eleven. Then since the parabola is normal width of one (one x squared term), then take the square root of the eleven height and get the width in either direction on the x-axis. So the parabola crosses the x-axis at about 0.683 and 7.317. If the parabola's x squared term was two, then the parabola would probably be skinnier since y gets twice as big. So in that case, you'd probably take the eleven(distance bottom of parabola is from x-axis) in this example and divide it by two before square rooting it. I just made this all up, so enjoy it.
Another way is by the "completing the square".
quadratic formula roxxors t3h big onez!111
At least I'm learning more... My math skills are pretty stiff... Thanks.
I think there is a simple way to do this
That's awesome... >_<;;;
Wow, lol, okay, never mind...
I'm sorry, but what formula was that? I'm not supposed to be knowing the answers before doing the problem... I am wondering how to determine the zeros by starting with the function itself; x^(2) - 8x + 5... Apparently my math skills are not top-notch... Thanks you so much, though...
Use the formula:
Hi, I am stuck with finding zeros for this one... >_<;; It may seem easy, but then I am confused with the answers I looked from the back of the book...