Question

Find the Domain and Range of xy - x + 3y =1

Answer 1

Given equation is xy-x+3y = 1

y(x+3)-x = 1

Add x to each side.

y(x+3)-x+x = 1+x

divide to each side by (x+3)

y(x+3)/(x+3) = (x+1)/(x+3)

Now the function is in the form y = f(x)

y = (x+1)/(x+3)

For x = 0 then y = 3/3 = 1

For x = -1 then y = 0/2 = 0

For x = 1 then y = 2/4 = 1/2

To find coordinate pairs substitute x = 0,1,-1 respectively then the set of coordinate pair would be y = 1,1/2,0.

We can able to write (0,1),(1,1/2),(-1,0).

We know that x values are domain and y values iis range of a function.

domain set is {0,-1,1}

Range set is {1,1/2,0}

Comments

Check your solution.........

Answer 2

The equation is xy - x + 3y = 1.

y (x + 3) = x + 1

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

So, the rational function is y = (x + 1)/(x + 3).

• We know that all possible values of x  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.

x + 3 ≠ 0

x ≠ - 3.

So, the domain of the function is all real numbers except - 3.(x ≠ - 3).

Domain : .

• The rational function is y = (x + 1)/(x + 3).

The graph of rational functions can be recognised by the fact two or more parts.

1).To find  y intercept, substitute x = 0 in the rational function.

y = (0 + 1)/(0 + 3)

⇒ y = 1/3.

The y - intercept is  1/3.

2). To find x - intercept, let the numarator = 0.

x + 1 = 0

⇒ x = - 1.

The x - intercept is  - 1.

3). The vertical asypototes are found  by solving the denominator = 0.

x + 3 = 0

⇒ x = - 3.

Vertical asymptotes is x = - 3.

4). To find horizontal asymptote.

Degree of the numarator = 1 and the degree of denominator = 1.

If the degree of the numerator is equal to the degree of the denominator,

leading coefficient of numarator is 1 and leading coefficient of denominator is also 1.

The horizontal asymptote is y = 1.

Now pick few more x - values, compute the corresponding y - values and flat few more points.

x	y = (x + 1)/(x + 3)	(x , y )
- 6	y = (- 6 + 1)/(- 6 + 3) = - 5/- 3 = 5/3	(- 6, 5/3)
- 4	y = (- 4 + 1)/(- 4 + 3) = - 3/- 1 = 3	(- 4, 3)
0	y = (0 + 1)/(0 + 3) = 1/3	(0, 1/3)
2	y = (2 + 1)/(2 + 3) = 3/5	(2, 3/5)
4	y = (4 + 1)/(4 + 3) = 5/7	(4, 5/7)

Graph :

1) Draw the coordinate plane.

2) Next dash the horizontal and vertical asympototes.

3) Plot the y intercept and coordinate pairs found in the table..

4) Connect the plotted points .

When, draw the graph, use smooth curves complete the graph.

From the graph range set is the corresponding values of the function for different values of x.

Range : .

Domain : , where R is the set of all real numbers.

Comments

There is a possibility to find the range without graphing (algebraically) as follows:

The function is xy - x + 3y =1 or y = (x + 1)/(x + 3).

Let the range be denoted R, which is also y - value, so we have xR - x + 3R =1 or R = (x + 1)/(x + 3).

Now, we want to solve for x :

xR - x + 3R =1

x(R - 1) + 3R =1

x(R - 1) = 1 - 3R

x = (1 - 3R)/(R - 1).

Now, x exist (and thus a correspondent range exist ) whenever R - 1 ≠ 0 → R ≠ 1.

Range : {y ∈ R : y ≠ 1}.

let the range be denoted R, which is also the y-value, so we have


Read more: http://www.physicsforums.com

let the range be denoted R, which is also the y-value, so we have


Read more: http://www.physicsforums.com

let the range be denoted R, which is also the y-value, so we have


Read more: http://www.physicsforums.com

let the range be denoted R, which is also the y-value, so we have


Read more: http://www.physicsforums.com

let the range be denoted R, which is also the y-value, so we have

Read more: http://www.physicsforums.com

let the range be denoted R, which is also the y-value, so we have

Read more: http://www.physicsforums.com