# Domain of (26x-42)/(x^2 -2x - 63) ?

• September 4th 2009, 07:29 PM
s3a
Domain of (26x-42)/(x^2 -2x - 63) ?
What is the domain of:

(26x-42)/(x^2 -2x - 63)

and how do I find it?

(My graphing software shows me something that I would have never thought of)

Any help would be greatly appreciated!
• September 4th 2009, 08:28 PM
Rapha
Hi s3a.

Quote:

Originally Posted by s3a
What is the domain of:

(26x-42)/(x^2 -2x - 63)

and how do I find it?

(My graphing software shows me something that I would have never thought of)

Any help would be greatly appreciated!

You need to find the zeros of the denominator
(x^2 -2x - 63) = 0

in this case the zeros are $x_1 = 9$ and $x_2 = -7$. You know how to solve (x^2 -2x - 63) = 0, right?

Therefore the denominator is $D = \mathbb{R} \backslash \{-7,9 \}$

Yours
Rapha
• September 4th 2009, 08:35 PM
dhiab
Quote:

Originally Posted by s3a
What is the domain of:

(26x-42)/(x^2 -2x - 63)

and how do I find it?

(My graphing software shows me something that I would have never thought of)

Any help would be greatly appreciated!

Hell0:
thie function is difinid $x^2 - 2x - 63 \ne 0
$

Solve : $x^2 - 2x - 63 = 0

$

$\left\{ \begin{array}{l}
a = 1,b = - 2,c = - 63 \\
\Delta = b^2 - 4ac \\
\end{array} \right.
$

$\Delta = 256$
$\begin{array}{l}
x_1 = \frac{{ - b + \sqrt \Delta }}{{2a}} \\
x_2 = \frac{{ - b - \sqrt \Delta }}{{2a}} \\
\end{array}
$

$\left\{ \begin{array}{l}
x_1 = 9 \\
x_2 = - 7 \\
\end{array} \right.
$

$D = \left] { - \infty , - 7} \right] \cup \left] { - 7,9} \right] \cup \left] {9, + \infty } \right]$