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 Pre-Calculus Pre-Calculus Math Forum

 August 14th, 2010, 03:10 PM #1 Newbie   Joined: Aug 2010 Posts: 2 Thanks: 0 PreCalculus Sorry, I wasn't too sure where to put this but this seemed right. Anyway, I'm stuck on a couple of questions and any help would be appreciated. 1. How can you sketch the graph of y = |x + 1| ? Translate the graph of y = |x| one unit _______. up, down, left, right. It's either up or right since it'll be positive, but I'm not too sure which one. 2. The graph of the equation y = x2 � 7 is symmetric with respect to which of the following? A.)the x-axis B.)the line 7-x = y C.)the y-axis D.) the line y = x I tried to graph the equation on a graphing calculator, but it said error and couldn't graph it. August 14th, 2010, 04:43 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: PreCalculus The first one will be horizontally translated one unit to the left. The 1 being added to x within the absolute value operator causes y to have the value it would for |x| only one unit smaller, or to the left. The second one I am assuming is y = x^2 - 7 and since it is the graph of y = x^2 translated 7 units down, would still have as its axis of symmetry, the y axis. August 14th, 2010, 04:49 PM   #3
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Re: PreCalculus

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 Originally Posted by MarkFL The first one will be horizontally translated one unit to the left. The 1 being added to x within the absolute value operator causes y to have the value it would for |x| only one unit smaller, or to the left. The second one I am assuming is y = x^2 - 7 and since it is the graph of y = x^2 translated 7 units down, would still have as its axis of symmetry, the y axis.
Thank you! That makes sense. And yes, I forgot to add the symbol to make it squared. August 14th, 2010, 04:57 PM #4 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: PreCalculus You're welcome!  August 14th, 2010, 05:21 PM #5 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: PreCalculus Hello Amanda, Some additional information on the first question: |a| = a als a >0 or a = 0 |a| = -a als a < 0 or a=0 If you would make a graph of "a", all the negative values for a would be reflected with respect to the x-axis to get |a|. Because of this, the graph of |a| will touch the x-axis. If |x| would go 1 down, |x+1| would have some negative values. 1 up would mean that |x+1| would not "touch" the x-axis. So for now it is whether to the left, or to the right. It is useful to know where x+1=0. This is when x=-1. The graph is afoot (x+1 increases whenever x gets larger, shifted to the right on a graph) it's slope is larger than 0. all values for x<-1 will be reflected with respect to the x-axis, so, when x=-1, the graph of |x+1| "touches" the x-axis. That is a translation of the graph of y = |x| one unit to the left. Hoempa Tags precalculus Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post magichands Academic Guidance 5 February 22nd, 2013 12:49 AM daivinhtran Algebra 1 September 18th, 2011 08:55 AM jkh1919 Algebra 3 September 5th, 2011 04:32 PM jjoiner Algebra 1 September 1st, 2011 06:25 AM jjoiner New Users 0 December 31st, 1969 04:00 PM

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