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.
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.
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.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?