anonymous
  • anonymous
Find the x-intercepts: 2(x-5)^2=17
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
x= [17/2]^1/2 +5|dw:1345790422320:dw|
anonymous
  • anonymous
x-intercepts? this is an equation in one variable.... how do you do that?
anonymous
  • anonymous
http://finedrafts.com/files/Larson%20PreCal%208th/Larson%20Precal%20CH2.pdf

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anonymous
  • anonymous
Am I suppose to distribute the 2?
anonymous
  • anonymous
ahh.... my bad.... because this is an equation in one variable, it is just a vertical line.... so solving for x will give you the only x-intercept....
anonymous
  • anonymous
since this is a quadratic function, you will get 2 values for the x-intercepts. read pp 126-129 in the link I've posted above.
anonymous
  • anonymous
2 x-intercepts? \[2(x-5)^2=17\]Divide both sides by 2\[(x-5)^2=\frac{17}{2}\]Take square roots for both sides \[(x-5)=\pm \frac{17}{2}\]Add 5 to both sides to get the answer.
anonymous
  • anonymous
I'm always late....
anonymous
  • anonymous
oh okay thank you that makes so much sense :)
anonymous
  • anonymous
hold on folks... this is an equation in x only.... this is a vertical line.... and not a function.... "Find the x-intercepts: 2(x-5)^2=17"
anonymous
  • anonymous
Wow. That's complicated. I thought all I had to do was to solve it but seems not :(
anonymous
  • anonymous
this is a function: y = 2(x-5)^2 this is not a function: 17 = 2(x-5)^2
anonymous
  • anonymous
parabolas always have 2 x-intercepts unless k=0, is it not?
anonymous
  • anonymous
But how to get the x-intercept(s)??? There is no y...
anonymous
  • anonymous
in any function, to find the x-intercepts, set y=0 and solve.... how are you gonna set y=0 in the equation 2(x-5)^2 =17 ????
razor99
  • razor99
Think the x-intercept is 8.5,0
anonymous
  • anonymous
Perhaps this would be the case? y = 2(x-5)^2-17 Put y=0 2(x-5)^2-17 = 0 2(x-5)^2=17 . . . Solve x to find the x-intercepts. Hmm...
anonymous
  • anonymous
dpalnc solved it, x=5 - sqrt(17/2) x=5+ sqrt(17/2)
anonymous
  • anonymous
compare these two equations: y = x + 3 and 17 = x + 3 that first one is a line and you can find the x-intercept by setting y=0 then solving 0 = x + 3. but that second equation is just a vertical line....
anonymous
  • anonymous
Eh?! Then, for 17=x+3, the x-intercept is 17-3 = 14?!
lgbasallote
  • lgbasallote
you have been trolled
anonymous
  • anonymous
dang this chrome browser.... x = 14 is a vertical line and that's where it crosses the x-axis.... but back to the problem... did i say 1 x-intercept? i mean two as panlac says.... :)
anonymous
  • anonymous
LOL
lgbasallote
  • lgbasallote
a vertical line that curves...obviously troll ^^
anonymous
  • anonymous
how 'bout two vertical lines... it's implied when u solve a quadratic you have to consider the positive and negative square root....
lgbasallote
  • lgbasallote
^trOWL
anonymous
  • anonymous
yeah buwahhahahahaha
anonymous
  • anonymous
ang troll ^^^
lgbasallote
  • lgbasallote
this goes against the teachings of the monks... the forefathers defined x-intercept as "the value of y when x is 0" however, there is no y....
lgbasallote
  • lgbasallote
this is not a function
anonymous
  • anonymous
right... so when you solve for x in that equation, you get two vertical lines... an equation in only 1 variable x is not a function...
lgbasallote
  • lgbasallote
no. this is just not a function. nothing more; nothing less
lgbasallote
  • lgbasallote
it's a relation, but not a function
lgbasallote
  • lgbasallote
x-intercepts occur in function only
anonymous
  • anonymous
when did i say it is a function?
anonymous
  • anonymous
I said it was, my mistake
lgbasallote
  • lgbasallote
you were saying it had an x-intercept
anonymous
  • anonymous
two values for x...
lgbasallote
  • lgbasallote
it has values for x...but no x-intercept
anonymous
  • anonymous
but the vertical line intercepts the x-axis...
anonymous
  • anonymous
yes. LG you are right
lgbasallote
  • lgbasallote
http://mathworld.wolfram.com/x-Intercept.html "The point at which a curve or FUNCTION crosses the x-axis (i.e., when in two dimensions)."
lgbasallote
  • lgbasallote
like i said. you got trolled
anonymous
  • anonymous
cross or touches...
anonymous
  • anonymous
so are you saying that a circle (which is a curve) does not have x-intercepts?
lgbasallote
  • lgbasallote
yes it doesnt. it's not a function
anonymous
  • anonymous
circle is not a function
lgbasallote
  • lgbasallote
neither do ellipses
anonymous
  • anonymous
function = 1 to 1 value
anonymous
  • anonymous
so why can't you call the where a vertical line crosses the the x-axis the x-intercept?
lgbasallote
  • lgbasallote
any closed figure is not a function. it is a plane
anonymous
  • anonymous
this is going beyond the problem now... bottom line: x has two values
anonymous
  • anonymous
i agree to agree... if that makes sense....
anonymous
  • anonymous
hey LG... i thnk u scared off the asker.... or he/she got bored...
anonymous
  • anonymous
did she ask the problem to be graphed?
anonymous
  • anonymous
lol jk. let us just leave it
lgbasallote
  • lgbasallote
i would like to quote myself "you got trolled" relations have x-intercepts too http://en.wikibooks.org/wiki/Algebra/Intercepts
anonymous
  • anonymous
yes.... let's graph it....
anonymous
  • anonymous
this is really going beyond now... why can't we agree that the intercepts are where either the vertical line or horizontal line are touched or crossed? did I start the fire when I said it was a function? I retracted it so OS can be a better place for students once again...
anonymous
  • anonymous
sorry man... i just miss talkin to LG....
anonymous
  • anonymous
u know u'd make a great ambassador for keeping the peace....:)
anonymous
  • anonymous
so peace to everyone.....:)
anonymous
  • anonymous
I'd be damned...
anonymous
  • anonymous
lol...

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