# What is the domain and range of y = sqrt(x-10) + 5?

Aug 3, 2015

Domain: $\left[10 , + \infty\right)$
Range: $\left[5 , + \infty\right)$

#### Explanation:

Let's start with the domain of the function.

The only restriction you have will depend on sqrt(x-10. Since the square root of a number will produce a real value only if that number if positive, you need $x$ to satisfy the condition

$\sqrt{x - 10} \ge 0$

which is equivalent to having

$x - 10 \ge 0 \implies x \ge 10$

This means that any value of $x$ that is smaller than $10$ will be excluded from the function's domain.

As a result, the domain will be $\left[10 , + \infty\right)$.

The range of the function will depend on the minimum value of the square root. Since $x$ cannot be smaller than $10$, f(10 will be the starting point of the function's range.

$f \left(10\right) = \sqrt{10 - 10} + 5 = 5$

For any $x > 10$, $f \left(x\right) > 5$ because $\sqrt{x - 10} > 0$.

Therefore, the range of the function is $\left[5 , + \infty\right)$

graph{sqrt(x-10) + 5 [-3.53, 24.95, -3.17, 11.07]}

SIDE NOTE Move the focus of the graph 5 points up and 10 points to the right of the origin to see function.