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!

Thanks in advance!

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- Sep 4th 2009, 07:29 PMs3aDomain 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!

Thanks in advance! - Sep 4th 2009, 08:28 PMRapha
Hi s3a.

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 - Sep 4th 2009, 08:35 PMdhiab
Hell0:

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.

$

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