# Help with Algebra 2 homework?

I need so much help with this! I have a horrible teacher, and no one else in my class has any idea what is going on. Any help would be great!

a. y=- √x+5-3

What is the:

1. Domain
2. Range
Of the equation above?

--------------------------------------…

b. y=(1/x-1)+3

What is the:

1. Domain
2. Range

Of the above equation?

Thanks for the help!!

+1 vote

b). y = 1/(x - 1) + 3

Let y = 0

0 = 1/(x -1) + 3

Rewrite the expression with common denominator.

0 = [1 + 3(x - 1)]/(x - 1)

Multiply each side by (x - 1).

1 + 3x - 3 = 0

3x - 2 = 0.

3x = 2

Divide each side by 3.

x = 2/3.

Domain: {x|x = 2/3} and Range: {y|y=0}

The equation

domain is

Range

a).  y = -√(x + 5) - 3------------>(1)

Let y = 0

-√(x + 5) - 3 = 0.

- √(x + 5) = 3

Multiply each side by negative one.

√(x + 5) = - 3

Take square each side.

x + 5 = 9.

Subtract 5 from each side.

x = 4.

Substituting x = 4 in the equation (1)

y = -√(4 + 5) - 3 = - √(9)-3 = - 3 - 3 = - 6

domain={x | x ≤ 4}, range = {y | - 6 ≤ y ≤ 0}.

The equation

Domain is { xR : x  ≥ 0 }.

Range is { yR : y  ≤ 2 }.

The equation

domain is { xR : x ≥ -5 } and range is  { yR : y  ≤ -3 }.

Domain: The domain of a function is the set of all possible input values (often the "x" variable), which produce a valid output from a particular function. It is the set of all real numbers for which a function is mathematically defined

Domain of a function f(x) is set of those values of x which will make the function mathematically legal or correct.. certain operations like division by zero , square root of a negative number do not exist in real maths.

So, the domain of the above function is all non negative real numbers.

The domain of is { x ∈ R : x ≥ 0 }.

Range: The range is the set of all possible output values (usually the variable y, or sometimes expressed as f(x)), which result from using a particular function.

Range set is the corresponding values of the function for different values of x.

Since for all non negative real numbers of x , the function is ≤ 2.

For example the first domain value is 0, then y = -(x)1/2  + 2= -(0)1/2+ 2= 2, so the range value starting with 2.

So, the range of the function is always less than or equal to 2.

Range

b)The function

• To determine the domain of the function

We know that all possible values of is domain of a function.

A rational function is simply fraction and in a fraction the denominator cannot be equal to 0 because it would be undefined.

To find which number make the fraction undefined create an equation where the denominator is not equal to zero.

So the domain of the function all real numbers except

• To determine the range ,solve for x

We know that all possible values of y is range of a function.

A rational function is simply fraction and in a fraction the denominator cannot be equal to 0 because it would be undefined.

To find which number make the fraction undefined create an equation where the denominator is not equal to zero.

Range

a) The equation

Domain: The domain of a function is the set of all possible input values (often the "x" variable), which produce a valid output from a particular function. It is the set of all real numbers for which a function is mathematically defined

Domain of a function f(x) is set of those values of x which will make the function mathematically legal or correct.. certain operations like division by zero , square root of a negative number do not exist in real maths.

So, the domain of the above function is all non negative real numbers.

The domain of is { xR : x ≥ -5 }.

Range: The range is the set of all possible output values (usually the variable y, or sometimes expressed as f(x)), which result from using a particular function.

Range set is the corresponding values of the function for different values of x.

Since for all non negative real numbers of x , the function is ≤ 3.

For example the first domain value is -5, then y = -(-5+5)1/2 - 3= -(0)1/2- 3 = -3, so the range value starting with -3.

So, the range of the function is always less than or equal to -3.

Range is { yR : y  ≤ -3 }.