y = (1/2)^(-1) = 1 / (1/2) = 2
y = (1/2)^(-2) = 1 / (1/2)² = 1 / (1/4) = 4
y = (1/2)^(-3) = 1 / (1/3)² = 1 / (1/9) = 9
y = 2 is a horizontal line thro (0,2)
y = 4 """"""""""""""""""""""(0,4)
y = 9"""""""""""""""""""""(0,9)
2007-09-30 20:06:35
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answer #1
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answered by Como 7
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ok so y = 1/2 is simply a constant so it is a line of y=1/2 at every x, so it is parallel to the x axis ok. so when you take 1/2 to the power or -1 it becomes y=2, then when you take it to the power of -2, it becomes y=4, then with -3 as a pwer it becomes, y= 8 and so on. so it is a bunch of horizontal lines,
if you have (1/2)^-1(or any negative number) what happens is the number that is being put to the negative power, you can takes its reciprical, which means for example, the reciprical of 8 is 1/8 the reciprical of 5 is 1/5, the reciprical of 1/2 is 2. so you take the reciprical of its positive power. so in the case of (1/2)^-3 you take 1/(1/2)^3 = 1/1/8 = 8
2007-10-01 00:35:49
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answer #2
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answered by LeonardoDaVinci 2
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Just graph y = 2^1, 2^2, 2^3 etc
It would be an exponentially rising curve
.
2007-10-01 00:27:03
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answer #3
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answered by tsr21 6
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Saying x is raised to -n is the same as saying the reciprocal of x raised to the n.
So, (1/2)^-1 = 1/(1/2) = 2
(1/2)^-2 = 1/(1/2)^2 = 1/(1/4) = 4
etc.
2007-10-01 00:26:17
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answer #4
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answered by jruzzy 2
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Remember, there is no x value in any of these lines - so you have a horizontal line!
y = (1/2)^(-1)
remember your index laws! x^(-a) = 1/(x^a)
so, 1/2 = 2^(-1)
y = (2^(-1))^(-1)
Index law: (a^x)^y = a^xy
so y = 2^1
the pattern continues with the negatives cancelling every time, but the number remaining. you will get:
for y = (1/2)^(-2)
y= 2^2
=4
next comes 2^3, 2^4, etc. All horizontal lines.
2007-10-01 00:32:12
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answer #5
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answered by mevelyn2551 3
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it is very easy.
the negative exponent means you should "flip" the fraction.
e.g. (1/2)^(-2) where " ^ " means "to the power of" is equal to (2)^(2) which is simply 4.
i'm sure you can graph y = 4.
2007-10-01 00:26:30
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answer #6
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answered by John H 4
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Please solve this math problem? It is very easy!!!?
It is very easy!!!?
assuming this is very easy as your claim says. then go on and do it.
2007-10-01 00:25:18
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answer #7
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answered by Anonymous
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well you said it was easy
so dooo it
2007-10-01 00:30:21
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answer #8
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answered by natalie 4
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