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