# Thread: The Graphs of Function and Relation

1. ## The Graphs of Function and Relation

Graph each equation and identify if it's a mere relation or function by vertical line test.

1.) |y|=x
2.) y= x raise to the 3rdpower + 1
3.) 2y = ( -2x+3 )

I DON'T know how to answer help pls.

2. ## re: The Graphs of Function and Relation

Originally Posted by ejaykasai
Graph each equation and identify if it's a mere relation or function by vertical line test.

1.) |y|=x
2.) y= x raise to the 3rdpower + 1
3.) 2y = ( -2x+3 )

I DON'T know how to answer help pls.
Dear ejaykasai,

Let me give you some hints;

For the first one use the definition of the absolute value function.

$\displaystyle |y| = \begin{cases} y & \mbox{if } y \ge 0 \\ -y & \mbox{if } y< 0. \end{cases}$

For the second equation use the fist and second derivatives to sketch the graph. If you do not know how to do this please refer Graphing Using First and Second Derivatives

The third one can be rearranged as, $\displaystyle 2y=-2x+3\Rightarrow{y=-x+\frac{3}{2}}$ So this is a equation of a straight line.

Hope you can continue.

3. ## Re: The Graphs of Function and Relation

Originally Posted by Sudharaka
Dear ejaykasai,

Let me give you some hints;

For the first one use the definition of the absolute value function.

$\displaystyle |y| = \begin{cases} y & \mbox{if } y \ge 0 \\ -y & \mbox{if } y< 0. \end{cases}$

For the second equation use the fist and second derivatives to sketch the graph. If you do not know how to do this please refer Graphing Using First and Second Derivatives

The third one can be rearranged as, $\displaystyle 2y=-2x+3\Rightarrow{y=-x+\frac{3}{2}}$ So this is a equation of a straight line.

Hope you can continue.
My problem is that I don't know how to remove the absolute value of y? " | y | "

4. ## Re: The Graphs of Function and Relation

What I would simply do is, since I know how to draw the graph of $\displaystyle y=|x|$, draw it and rotate the whole graph 90 degrees so that y and x axis swap there places

5. ## Re: The Graphs of Function and Relation

Originally Posted by ejaykasai
My problem is that I don't know how to remove the absolute value of y? " | y | "
That is exactly what you could do with the definition of the absolute value function.

$\displaystyle |y| = \begin{cases} y & \mbox{if } y \ge 0 \\ -y & \mbox{if } y< 0. \end{cases}$

$\displaystyle |y|=x$

Therefore, $\displaystyle y=x\mbox{ when }y\geq 0$----------(1)

$\displaystyle -y=x\mbox{ when }y<0\Rightarrow y=-x\mbox{ when }y<0$-------(2)

Combining (1) and (2) we can write;

$\displaystyle y= \begin{cases} x & \mbox{if } y \ge 0 \\ -x & \mbox{if } y< 0. \end{cases}$

Hope you can sketch the graph now.

6. ## Re: The Graphs of Function and Relation

What I would simply do is, since I know how to draw the graph of $\displaystyle y=|x|$, draw it and rotate the whole graph 90 degrees so that y and x axis swap there places
Technically, the operation you need to do is reflect the graph about the line y = x. That happens to do the same thing in this case, but in general, the rotation is incorrect. Any time you swap the roles of x and y, you are graphically reflecting about the line y = x.

7. ## Re: The Graphs of Function and Relation

Is this right?
instead of assigning first values of x, values of y 1st, then solve for x:

Example:

when y = 0, x = 0 (0, 0)
when y = 1, x = 1 (1, 1)
when y = -1, x = 1 (1, -1)
when y = 2, x = 2 (2, 2)
when y = -2, x = 2 (2, -2)

when I solved the |y| = x using the default values, the graph goes like this-->
|y|=x | Flickr - Photo Sharing! ?? Is this correct?

8. ## Re: The Graphs of Function and Relation

Is this right?
instead of assigning first values of x, values of y 1st, then solve for x:

Example:

when y = 0, x = 0 (0, 0)
when y = 1, x = 1 (1, 1)
when y = -1, x = 1 (1, -1)
when y = 2, x = 2 (2, 2)
when y = -2, x = 2 (2, -2)

when I solved the |y| = x using the default values, the graph goes like this-->
|y|=x | Flickr - Photo Sharing! ?? Is this correct?
No your graph is incorrect. Please check whether the highlighted values are marked in the graph correctly.