tatianagomezb
  • tatianagomezb
Hi I would like help solving this exercise: Consider the function f (x)= |x^2 -6x+3| a- graph f I know what absolute value means, and I found the solution to x^2-6x+3= 0; it's 5 and 1. But I don't understand how to graph it. I know that |x|= x if x> 0 and -x if x <0. But how can I translate it to the graph. Thanks
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You can graph it by putting it in a CAS program. When it is in | |, then it only spits positive values on the y axis. http://www.wolframalpha.com/input/?i=f(x)%3D%7Cx%5E2-6x%2B3%7C
tatianagomezb
  • tatianagomezb
Thanks for the tip I have a similar software. The thing is I'm studying for precalc finals and I have to understand why the graph turns out that way, because after I graph it there are several questions to be answered. Like: When does the function has 4 different y values ?
anonymous
  • anonymous
If you can graph \(x^{2}-6x+3\) Normally, then you can easily graph the absolute value of it. Are you comfortable with graphing \(x^{2}-6x+3?\)

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anonymous
  • anonymous
@tatianagomezb
anonymous
  • anonymous
And if you found solutions to be 5 and 1, then your function should be \(x^{2}-6x+5\) or some multiple of it.
tatianagomezb
  • tatianagomezb
Yes
anonymous
  • anonymous
Did you mean \(x^{2}-6x+5?\)
tatianagomezb
  • tatianagomezb
Hold on
tatianagomezb
  • tatianagomezb
Yes, typo. Yes it is x^2 -6x +5
anonymous
  • anonymous
Okay. So would you be comfortable enough using the draw tool on here to graph \(x^{2}-6x+5?\) From there I can easily show you how to get the absolute value version.
tatianagomezb
  • tatianagomezb
Okay,
tatianagomezb
  • tatianagomezb
Done
anonymous
  • anonymous
Did the graph get posted? I don't see that you posted the graph, lol.
tatianagomezb
  • tatianagomezb
I'm on my phone, I don't rleally know Here http://www.cymath.com/answer.php?q=graph%20y%3Dx%5E2-6x%2B5
anonymous
  • anonymous
Oh, I see. That complicates using the draw tool then, lol. Alright, Ill draw it out and show ya.
tatianagomezb
  • tatianagomezb
I wanted to upload the screen shot but I couldn't
anonymous
  • anonymous
|dw:1449821020376:dw| Bad graph, but whatever. Still will show the idea.
anonymous
  • anonymous
Refer to the attached plot from Mathematica.
1 Attachment
anonymous
  • anonymous
So you know the definition of absolute value, but that extends to a more general idea |dw:1449821155090:dw| As in, for \(x^{2}-6x+5\), the absolute value of it means this: |dw:1449821278322:dw| So if you can find out when the function is greater than or equal to 0 and less than 0, you can graph it by plotting the original function for all values of x in which the function is positive. For the x-values that make the function negative, graph the same function but multiplied by -1.
tatianagomezb
  • tatianagomezb
Thanks @concentrationalizing and @robtobey. Okay, that's where I was going. When the absolute value is positive then what? Is the graph going to be above 0, that is the part I'm confused by
anonymous
  • anonymous
Now in regards to graphing, the idea is even simpler than that. I graphed the original function without absolute value because all I need to do from there to graph the absolute value version is to take the part below the x-axis and flip it upwards like this: |dw:1449821589225:dw| I took only the part that dipped below and flipped it up and mirrored it across the x-axis.
anonymous
  • anonymous
For another quick example, let me just graph an arbitrary function. |dw:1449821707716:dw| I have no idea what this function is (probably some sort of -x^4 + ... function), but regardless of what the function is, if I want the graph of its absolute value, I just take all the parts that are below the x-axis and reflect them upward
anonymous
  • anonymous
|dw:1449821782283:dw| And thats it. The absolute value is the same thing, but with the negative parts mirrored upward. So in general, to do the absolute value graphs, try to graph the function without absolute value and then make the parts below the x-axis flip upward.
tatianagomezb
  • tatianagomezb
Okay, I guess that's better. Sometimes math gets so complicated. Thanks again guys!
anonymous
  • anonymous
No problem :)

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