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How do you find the domain and range of #y = sqrt(x+8)#?

Answer 1

Eli Assouline

Domain is #x >= -8#

Range is #y >= 0#

First, let's examine the equation.

# y = sqrt(x+8)#

We are interested in the values of x for the domain that result in a "valid result".

Alternatively put, we're searching for x values that won't "break" the equation.

We notice that #x+8# is inside of a square root. We also know that anything inside of a square root must be non-negative, since you can't take the square root of a negative number.

We can therefore set this up.

#x+8 >=0#

To obtain, subtract 8 from each side.

#x >= -8# , which is the domain.

Let's talk about the range now.

We know that the square root of a number cannot be negative. This is the same for function such as #x + 8#. So, we get:

#y >= 0#

It's important to remember that the square root symbol only denotes the positive root in this case. For instance,

#sqrt(4) = 2#

while if #x^2 = 4#, then #x = +-2#

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Answer 2

Eva Abramson

The domain of y = √(x + 8) is x ≥ -8, and the range is y ≥ 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.