Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
5. Graphing Square Root Functions
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Exercise 6 Page 641

Think about the domain of a radical function that has a difference of a variable and some number under the square root. Keep in mind that it is not possible to calculate a square root of a negative number.

Yes, see solution.

Practice makes perfect
Let's start by recalling that a square root function is a function containing a square root with the independent variable in the radicand. The parent square root function is the following. y=sqrt(x) Now, let's think about what values x can take. The square root can only be calculated for a number that is greater than or equal to 0 — we cannot calculate the square root of a negative number.

Positive:& sqrt(2), sqrt(5), sqrt(81) ✓ Negative:& sqrt(- 2), sqrt(- 8), sqrt(- 19) * The set of values of the independent variable is called a domain. We conclude that the domain of y=sqrt(x) is limited to x≥ 0. Now, let's find the domains of the other two square root functions. y=sqrt(x+5) and y=sqrt(x-3) In order to do this, we will use the fact that the radicand of a square root function should be a non-negative number. We can form two inequalities to find these domains.

Function y=sqrt(x+5) y=sqrt(x-3)
Inequality x+5≥ 0 x-3≥ 0
Solution x≥ - 5 x≥ 3
As we can see, in the first function, the value of x must be greater than or equal to -5. This means that the domain of this function includes negative values. However, the domain of the second function contains only positive values.
Therefore, we can conclude that if the parent square root function is translated to the left, it can have negative values in the domain.