How do you find the domain and range of y=(1/x)-5?

- anonymous

How do you find the domain and range of y=(1/x)-5?

- katieb

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- anonymous

I already graphed.

- anonymous

|dw:1359707995799:dw|

- anonymous

notice the vertical asymptote... at x = ???
this should give you a hint for the domain...
notice the horizontal asymptote... at y = ???
this should give you a hint for the range.

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## More answers

- skullpatrol

What happens at x = 0?

- jim_thompson5910

Hint: To find the domain, find all the values of x that make the denominator equal to zero
To find the range, find the inverse function and repeat the same steps to find the domain

- anonymous

I'm confused with this one because there are two lines. It throws me off. Should I put them together or what?

- skullpatrol

What do the "lines" mean?

- anonymous

I'm not really sure what they mean, they're just lines

- anonymous

The two curve lines

- skullpatrol

|dw:1359708410347:dw|

- skullpatrol

What is the domain and range of this equation?

- anonymous

This is what I'm having trouble with. I don't know how to find the domain and the range for these kinds of functions

- skullpatrol

The domain is all the values that you are allowed to input into the equation.

- jim_thompson5910

ignore the graph
look at the equation y = (1/x) - 5
x cannot be 0 because if it were, then you would be dividing by zero (which is not allowed)
so you must restrict 0 from the domain
any other number is just fine
So the domain is the set of all real numbers but x cannot be zero

- jim_thompson5910

the range can be found in a similar way, but you first need to find the inverse function

- skullpatrol

What would those be for $$y=\frac{1}{x}$$

- anonymous

What do you do after you find the inverse function?

- jim_thompson5910

what did you get for the inverse function

- anonymous

Actually, nevermind.

- anonymous

I want to be able to look at the graph and know the domain and the range

- jim_thompson5910

I'll show you how to get the inverse function
Swap x and y, then solve for y
y = (1/x) - 5
x = (1/y) - 5
x+5 = 1/y
y(x+5) = 1
y = 1/(x+5)
So the inverse function is y = 1/(x+5)

- jim_thompson5910

what makes the denominator x+5 equal to zero ?

- jim_thompson5910

we can go back to the graph after we do these equations

- anonymous

Wait, hold on. @skullpatrol is it |dw:1359709153164:dw|

- skullpatrol

Yes, and why?

- skullpatrol

|dw:1359709317420:dw|

- skullpatrol

|dw:1359709670237:dw|

- skullpatrol

Compare your graph to mine :)

- anonymous

That's what I got

- anonymous

x=/=0
y=/=-5

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