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August 14th, 2010, 03:10 PM   #1
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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.
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August 14th, 2010, 04:43 PM   #2
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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.
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August 14th, 2010, 04:49 PM   #3
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Re: PreCalculus

Quote:
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.
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August 14th, 2010, 04:57 PM   #4
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Re: PreCalculus

You're welcome!
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August 14th, 2010, 05:21 PM   #5
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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.

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