I need help

(x+6) /(sqrt(x^2-3x-4) )

Has a domain for ALL x such that the denominator does NOT equal Zero..AND we do not have a NEGATIVE sqrt

let's find what those values are

sqrt(x^2-3x-4)  = sqrt(x-4)(x+1)       This WILL equal zero when x = 4 or -1 <------ NOT allowed !

ALSO from    - infinity to +4 is invalid as a negative sqrt will result

to NET it all out , the domain is         4< x  thru  + infinity

$f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}$

Note that     x^2 - 3x - 4    must be >  0

Set  this to 0

x^2  -3x - 4  = 0

(x - 4) ( x + 1)  = 0

Setting each factor to  0   and solving for x gives us

x = -1    and x  = 4

We have three  possible intervals

(-inf , -1)  U (-1, 4)   U (4, inf)

Testing x  = 0   in the middle interval makes the polynomial negative....so....this interval does not work

The domain is the other two intervals ⇒    (-inf, -1)  U  ( 4, inf )

Here's a graph  :   https://www.desmos.com/calculator/axfthxpp5t