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# How do you find the range of a function algebraically y=(x+5)/(x-2)?

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#### Explanation

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#### Explanation:

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120
Apr 13, 2015

To best way to find the range of a function is to find the domain of the inverse function. To find the inverse function of a function you have to substitue $x$ with $y$, and vice versa, and then find $y$.

So:

$y = \frac{x + 5}{x - 2} \Rightarrow x = \frac{y + 5}{y - 2} \Rightarrow x \left(y - 2\right) = y + 5 \Rightarrow$

$x y - 2 x = y + 5 \Rightarrow x y - y = 2 x + 5 \Rightarrow y \left(x - 1\right) = 2 x + 5 \Rightarrow$

$y = \frac{2 x + 5}{x - 1}$ and its domain is $\left(- \infty , 1\right) \cup \left(1 , + \infty\right)$ that is, also, the range of your function.

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

25
mason m Share
Jan 2, 2016

$\left(- \infty , 1\right) \cup \left(1 , + \infty\right)$

#### Explanation:

There will be a break in the range at any horizontal asymptote.

Since the degree of the numerator and denominator are equal, take and divide the coefficients of the terms with the largest degree.

Both of these terms are $x$, and $\frac{1}{1} = 1$. Thus, there is a horizontal asymptote at $y = 1$.

The range is $\left(- \infty , 1\right) \cup \left(1 , + \infty\right)$.

graph{((x+5)/(x-2)-y)(y-1)=0 [-28.45, 29.27, -12.3, 16.57]}

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