What is the domain of y=(sqrt(x^2+4x))/(2x+6) ?

The domain of y=(sqrt(x^2+4x))/(2x+6) is x = 0

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domain of y=(sqrt(x^2+4x))/(2x+6)

domain

y = \frac{x ^{2} + 4 x}{2 x + 6}

x \leq - 4 or x \geq 0

Interval Notation

(- \infty, - 4] \cup [0, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 4 or x \geq 0

Find undefined (singularity) points:

x = - 3

Combine real regions and undefined points for final function domain

x \leq - 4 or x \geq 0

g (x) = (2 x)^{2} - 3

\frac{x + 1}{x - 1}

2 - x^{2}

f (x) = \frac{x + 2}{x ^{2} - 6 x + 9}

What is the domain of y=(sqrt(x^2+4x))/(2x+6) ?
The domain of y=(sqrt(x^2+4x))/(2x+6) is x<=-4\lor x>= 0