# What is the domain and range of #y=-sqrt(x^2-3x-10)#?

##### 1 Answer

#### Answer:

Domain:

Range:

#### Explanation:

Right from the start, you know that the expression under the square root must be **positive**, because for real numbers you can only take the square root of positive numbers.

Make the expression inside the square root equal to zero to determine its roots

#x^2 - 3x - 10 = 0#

#x_(1,2) = (-(-3) + - sqrt((-3)^2 - 4 * 1 * (-10)))/(2 * 1)#

#x_(1,2) = (3 +- sqrt(49))/2#

#x_(1,2) = (3 +- 7)/2 = {(x_1 = (3 + 7)/2 = 5), (x_2 = (3 - 7)/2 = -2) :}#

The quadratic can be factored using its two roots

#(x-5)(x+2) = 0#

In order for this to be greater than or equal to zero, you need either both *positive* or both to be *negative*. Do not forget to include the values that makes each term equal to zero.

For any value of

#{(x-5>=0), (x+2 > 0) :} implies (x-5)(x+2)>=0#

Likewise, for any value of

#{(x-5<0), (x+2 <= 0) :} implies (x-5)(x+2)>=0#

The domain of the original function will thus be

To find the range of the function, keep in mind that the square root of any positive real number is always **positive**. Since you have

#f(x) = - sqrt(x^2 - 3x - 10)#

you get that

#f(x) <= 0", "(AA)x in (-oo, -2] uu [5, + oo)#

The range of the function will thus be

graph{-sqrt(x^2 -3x - 10) [-10, 10, -5, 5]}