What is the domain and range of #y =sqrt(17x+8)#?

1 Answer

Answer:

Domain: #x>=-8/17# or Domain:#[-8/17, +oo)#
Range:#y>=0# or Range:#[0, +oo)#

Explanation:

The square root of a negative number is an imaginary number.
The square root of zero is zero. The radicand is zero at #x=-8/17#.
Any value greater than -8/17 will result to a positive radicand.
Therefore, Domain: #x>=-8/17#

Range: is 0 to +infinity

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