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 August 8th, 2009, 02:30 AM #1 Newbie   Joined: Aug 2009 Posts: 2 Thanks: 0 Parabola here is the image that came with the question...although there are 5 options none of them according to graph seem correct. and the answer they gave doesnt seem to add up either? the question and answer choices were: Which of the following could be the equation of the function graphed in the xy-plane above? a) y = (–x)2 + 1 b) y = –x2 + 1 c) y = |x2 + 1| d) y = |x2 – 1| e) y = |(x – 1)2| and the answer they gave was this: The function with equation y = (–x)2 + 1 and the function with equation y = |x2 + 1| each have a minimum value of 1 when x = 0, but the function graphed does not have a minimum value of 1, so these options cannot be correct. The graph of the function with equation y = –x2 + 1 contains the point (2, –3), but the function graphed does not contain any points with negative y-coordinates, so this option cannot be correct. The graph of the function with equation y = |(x – 1)2| is not symmetric with respect to the y-axis, so it cannot be the equation of the function graphed. Therefore, the only equation that could correspond to the function graphed is y = |x2 – 1|. Its graph is the absolute value of a parabola opening upward with vertex at (0, –1). please explain according to image there is no parabola in y axis pointing -1?
 August 8th, 2009, 03:33 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,974 Thanks: 1850 The supplied solution did not state that the graph is a parabola with vertex at (0, -1); it stated that the graph is the absolute value of a parabola opening upward with vertex at (0, –1). The wording "absolute value" tells you that negative values are replaced by the correponding positive values, which explains what you see in the image.
 August 8th, 2009, 12:48 PM #3 Newbie   Joined: Aug 2009 Posts: 2 Thanks: 0 Re: Parabola thanx
 August 8th, 2009, 03:56 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Since this is posted under Calculus: limits, differentiation, integration section, here are some calculus facts to know. Let f(x) be such an function. Let the equation y = f(x) = |x² – 1| be true. It is obvious that f'(±1) is undefined, f'(x) > 0 for -1 < x < 0 and x > 1, f'(x) < 0 for x < -1 and 0 < x < 1, f'(x) = 0 for x = 0, and we can also determine the concavity for x < -1, -1 < x < 1 and x > 1 using the second derivative of the function f(x).

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