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

1. ## 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!

2. Hi s3a.

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 $\displaystyle x_1 = 9$ and $\displaystyle x_2 = -7$. You know how to solve (x^2 -2x - 63) = 0, right?

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

Yours
Rapha

3. 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!
thie function is difinid $\displaystyle x^2 - 2x - 63 \ne 0$
Solve : $\displaystyle x^2 - 2x - 63 = 0$
$\displaystyle \left\{ \begin{array}{l} a = 1,b = - 2,c = - 63 \\ \Delta = b^2 - 4ac \\ \end{array} \right.$
$\displaystyle \Delta = 256$
$\displaystyle \begin{array}{l} x_1 = \frac{{ - b + \sqrt \Delta }}{{2a}} \\ x_2 = \frac{{ - b - \sqrt \Delta }}{{2a}} \\ \end{array}$
$\displaystyle \left\{ \begin{array}{l} x_1 = 9 \\ x_2 = - 7 \\ \end{array} \right.$
$\displaystyle D = \left] { - \infty , - 7} \right] \cup \left] { - 7,9} \right] \cup \left] {9, + \infty } \right]$