Well, the vertex is going to be the point of lowest value for this function. Since the absolute value of any number other than zero is greater than zero, we're looking for the spot where we're taking the absolute value of zero. Setting this up:
abs(x-2) = 0
Since zero is neither positive or negative, we get:
x - 2 = 0
x = 2
Now substitute that value into the original equation:
Answers & Comments
Verified answer
It's where the contents of the | | signs give the least amount … ie, where x = 2
y = |x - 2| - 5
y = |0| - 5
y = -5
As an ordered pair: (2, -5)
I can't plot the graph here on Y!A, but the following table illustrates the point:
x ..… y
––––––
-1 … -2
0 … -3
1 … -4
2 … -5 <<< vertex
3 … -4
4 … -3
5 … -2
- - - - - - - - - - - - - - - - - - - - - - - - - - -
<edit> Here's the WolframAlpha page: http://www.wolframalpha.com/input/?i=y+%3D+%7Cx+%E...
- - - - - - - - - - - - - - - - - - - - - - - - - - -
Well, the vertex is going to be the point of lowest value for this function. Since the absolute value of any number other than zero is greater than zero, we're looking for the spot where we're taking the absolute value of zero. Setting this up:
abs(x-2) = 0
Since zero is neither positive or negative, we get:
x - 2 = 0
x = 2
Now substitute that value into the original equation:
y = abs(2 - 2) - 5
y = 0 - 5
y = -5
Our vertex will occur at (2, -5)