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Can someone help me figure this question on functions and using transformations?Use...
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High School Teacher
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Given `f(x)=2^(-x)-2` :
(a) The domain is all real numbers. Exponentials accept any x as input.
(b) The range is y>-2. The value of `2^(-x)` goes to zero as x tends to infinity, so the limit of the function is -2.
Thinking in terms of transformations, we translate the graph of `y=2^(-x)` down 2 units. Since `y=2^(-x)` has a horizontal asymptote of y=0, the new horizontal asymptote is y=-2.
(c) The horizontal asymptote is y=-2.
(d) The base graph is `y=2^x` . Changing the x to -x results in a reflection across the y-axis, and subtracting 2 is a vertical translation down 2 units.
`y=2^x` in black;`y=2^(-x)` in green;`y=2^(-x)-2` in red.
Posted by embizze on December 16, 2011 at 11:09 AM (Answer #1)
oops, I forgot to give the function, that is f(x) =( 2^-2) - 2
Posted by deekoch on December 13, 2011 at 8:51 AM (Answer #2)
The question is a bit confusing, could you please rewrite the question?
Posted by rock-star790 on December 13, 2011 at 9:04 AM (Answer #3)
Is the question f(x)= (x^2-2) - 2?
Posted by rock-star790 on December 13, 2011 at 9:06 AM (Answer #4)
My question is: Use transformations to graph the following function and submit a graph. Also state a) the domain, b) the range, c) the horizontal asymptote. f(x) = (2^-x) - 2. The -x is the exponent in the question.
Posted by deekoch on December 14, 2011 at 8:18 AM (Answer #5)
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