Graphing functions

bandlover24

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Oct 11, 2011
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I know how to graph functions such as this one: f(x)= -1/x-3 -1 but I'm not sure how to determine the domain, range, and vertex in this particular problem. I'm confused because it's a radical. I don't have any work to show, sorry.
 
It is often easiest to assume "All Real Numbers" and then determine if some subset should be excluded.

I'm not really sure what you are working on. "f(x)= -1/x-3 -1" Is that \(\displaystyle f(x) = \frac{-1}{x-3}-1\)?

If so, please remember your Order of Operations and use parentheses where necessary.

Does this work for x = 3? Why or why not? If it doesn't work, it's not in the Domain.
 
It is often easiest to assume "All Real Numbers" and then determine if some subset should be excluded.

I'm not really sure what you are working on. "f(x)= -1/x-3 -1" Is that \(\displaystyle f(x) = \frac{-1}{x-3}-1\)?

The base graph is \(\displaystyle f(x) \ = \ \frac{1}{x}.\)

The (-1) in the numerator reflects it across the x-axis. The (x - 3) in the
denominator verticvally shifts it 3 units to the right. And the subtraction
of 1 after the fraction makes it have a vertical shift of down unit.

The base function has a horizontal asymptote of y = 0, as well as it does not
intersect with its horizontal asymptote (in this particular case.)

When the graph of the base function is relected across the x-axis, the horizontal
asymptote, y = 0, does not change. But when the base function is vertically
shifted 1 unit down, the horizontal asymptote is transformed to y = -1
for the function given. It is still the case that the curve does not cross its
horizontal asymptote (in this case).


For the range, it is:


\(\displaystyle (-\infty, -1) \cup (-1, \infty) \ \ or\)


\(\displaystyle \{y \ | \ y \ne -1\}\)
 
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