Math Help - Graphing a Absolute Value Equation

1. Graphing a Absolute Value Equation

I would like to know where my math went wrong with the following equation:
$y = 2|x-3| + 1$

So, to graph, I assumed on finding the x-intercept and y-intercept.
Set y = 0
$0 = 2|x-3| + 1$
$-1 = 2|x-3|$
$-1/2 = |x-3|$
$-1/2 = x-3$ and $-1/2 = -(x-3)$
$6/2 -1/2 = x$ and $-6/2 - 1/2 = -x$
$5/2 = x$ and $7/2 = x$

According to my math:
(5/2, 0) and (7/2, 0) are x-intercepts

But, when I check my math, $y>= 1$

What did I do wrong?

2. Re: Graphing a Absolute Value Equation

Originally Posted by litlmike
I would like to know where my math went wrong with the following equation:
$y = 2|x-3| + 1$

So, to graph, I assumed on finding the x-intercept and y-intercept.
Set y = 0
$0 = 2|x-3| + 1$
$-1 = 2|x-3|$ Mr F says: How can the right hand side possibly be equal to a negative number!? It should be obvious that there is no solution to this equation.

$-1/2 = |x-3|$
$-1/2 = x-3$ and $-1/2 = -(x-3)$
$6/2 -1/2 = x$ and $-6/2 - 1/2 = -x$
$5/2 = x$ and $7/2 = x$

According to my math:
(5/2, 0) and (7/2, 0) are x-intercepts

But, when I check my math, $y>= 1$

What did I do wrong?
The salient point is located at (3, 1) and the y-intercept is at (0, 7). That should be enough to correctly draw the graph.

3. Re: Graphing a Absolute Value Equation

Originally Posted by mr fantastic
The salient point is located at (3, 1) and the y-intercept is at (0, 7). That should be enough to correctly draw the graph.
Thanks for the feedback. So there is no step that is incorrect in the math steps? It is just that I am supposed to to recognize that it should be 'No Solution'? What if it were a much more difficult equation to recognize this truth, something with large number and a larger polynomial, how could I determine it then?

4. Re: Graphing a Absolute Value Equation

Originally Posted by litlmike
Thanks for the feedback. So there is no step that is incorrect in the math steps? It is just that I am supposed to to recognize that it should be 'No Solution'? What if it were a much more difficult equation to recognize this truth, something with large number and a larger polynomial, how could I determine it then?
See below for the mistake. As for polynomials and so on, you should be aware that because of the fundamental theorem of algebra, a polynomial of degree n has n complex solutions, counting multiplicities. What makes this equation problematic isn't the polynomial of degree one, but the absolute value signs. The absolute value of any complex number is a real number greater than or equal to zero.

Originally Posted by litlmike
I would like to know where my math went wrong with the following equation:
$y = 2|x-3| + 1$

So, to graph, I assumed on finding the x-intercept and y-intercept.
Set y = 0
$0 = 2|x-3| + 1$
$-1 = 2|x-3|$
$-1/2 = |x-3|$
Going from the preceding step to the subsequent step is the problem. The absolute value signs don't work this way. What you can say is that -1/2 = x-3 if x-3 is nonnegative, or -1/2 = -(x-3) if x-3 is negative. Still, in neither case can you possibly have solutions.

$-1/2 = x-3$ and $-1/2 = -(x-3)$
$6/2 -1/2 = x$ and $-6/2 - 1/2 = -x$
$5/2 = x$ and $7/2 = x$

According to my math:
(5/2, 0) and (7/2, 0) are x-intercepts

But, when I check my math, $y>= 1$

What did I do wrong?

5. Re: Graphing a Absolute Value Equation

Hello, litlmike!

I would like to know where my math went wrong with the following equation:
. . $y \:=\: 2|x-3| + 1$

So, to graph, I assumed on finding the x-intercept and y-intercept: set $y = 0$

. . $2|x-3| + 1 \:=\:0 \quad\Rightarrow\quad 2|x-3| \:=\:-1$

. . $|x-3| \:=\:-\tfrac{1}{2}$ . Here!

An absolute value is always nonnegative (greater than or equal to zero).

So your equation has no solutions . . . There are no x-intercpts.

You can sketch the graph using "transformations" . . .

We know the graph of $y \:=\:|x|$
. . It is a V-shape with its vertex at the origin.

Code:
              |
*       |       *
*     |     *
*   |   *
* | *
- - - - - o - - - - - -
|

The graph of: $y \:=\:|x-3|$ is the same graph moved 3 units to the right.

Code:
              |
* |             *
*           *
| *       *
|   *   *
- - - - + - - o - - - - -
|     3

The graph of: $y \:=\:2|x-3|$ is the previous graph except the sides are "steeper".

Code:
              |
| *       *
|  *     *
|   *   *
|    * *
- - - - + - - o - - - - -
|     3

The graph of: $y \:=\:2|x-3|+1$ is the previous graph moved up one unit.

Code:
              |
| *       *
|  *     *
|   *   *
|    * *
|     o(3,1)
|
- - - - + - - - - - - -
|     3
. . There!

6. Re: Graphing a Absolute Value Equation

Originally Posted by litlmike
I would like to know where my math went wrong with the following equation:
$y = 2|x-3| + 1$
If x< 3, x-3< 0 so |x-3|= -(x- 3)= 3- x. If x< 3, y= 2(3- x)+ 1= 7- 2x. That is a straight line with slope -2, passing through (0, 7) and ending at (3, 1).

If $x\ge 3$, x- 3> 0 so |x-3|= x- 3. If $x\ge 3$, y= 2(x- 3)+ 1= 2x- 5. That is a straight line with slope 2, connecting to the previous line at (3, 1).

7. Re: Graphing a Absolute Value Equation

$\displaystyle y = x$

$\displaystyle y = x - 3$ is a vertical translation down by $\displaystyle 3$ units

$\displaystyle y = |x - 3|$ is a reflection of everything negative in the $\displaystyle x$ axis

$\displaystyle y = 2|x - 3|$ is a dilation of factor $\displaystyle 2$

$\displaystyle y = 2|x - 3| + 1$ is a vertical translation of $\displaystyle 1$ unit up.

8. Re: Graphing a Absolute Value Equation

Too many cooks spoil the broth. If anyone other than litlmike responds to this thread next, I am going to infract.