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anonymous
 4 years ago
Find the xintercepts: 2(x5)^2=17
anonymous
 4 years ago
Find the xintercepts: 2(x5)^2=17

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x= [17/2]^1/2 +5dw:1345790422320:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0xintercepts? this is an equation in one variable.... how do you do that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://finedrafts.com/files/Larson%20PreCal%208th/Larson%20Precal%20CH2.pdf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Am I suppose to distribute the 2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh.... 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 xintercept....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since this is a quadratic function, you will get 2 values for the xintercepts. read pp 126129 in the link I've posted above.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02 xintercepts? \[2(x5)^2=17\]Divide both sides by 2\[(x5)^2=\frac{17}{2}\]Take square roots for both sides \[(x5)=\pm \frac{17}{2}\]Add 5 to both sides to get the answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh okay thank you that makes so much sense :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hold on folks... this is an equation in x only.... this is a vertical line.... and not a function.... "Find the xintercepts: 2(x5)^2=17"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wow. That's complicated. I thought all I had to do was to solve it but seems not :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is a function: y = 2(x5)^2 this is not a function: 17 = 2(x5)^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0parabolas always have 2 xintercepts unless k=0, is it not?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But how to get the xintercept(s)??? There is no y...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in any function, to find the xintercepts, set y=0 and solve.... how are you gonna set y=0 in the equation 2(x5)^2 =17 ????

razor99
 4 years ago
Best ResponseYou've already chosen the best response.0Think the xintercept is 8.5,0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Perhaps this would be the case? y = 2(x5)^217 Put y=0 2(x5)^217 = 0 2(x5)^2=17 . . . Solve x to find the xintercepts. Hmm...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dpalnc solved it, x=5  sqrt(17/2) x=5+ sqrt(17/2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0compare these two equations: y = x + 3 and 17 = x + 3 that first one is a line and you can find the xintercept by setting y=0 then solving 0 = x + 3. but that second equation is just a vertical line....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Eh?! Then, for 17=x+3, the xintercept is 173 = 14?!

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0you have been trolled

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dang this chrome browser.... x = 14 is a vertical line and that's where it crosses the xaxis.... but back to the problem... did i say 1 xintercept? i mean two as panlac says.... :)

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0a vertical line that curves...obviously troll ^^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how 'bout two vertical lines... it's implied when u solve a quadratic you have to consider the positive and negative square root....

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0this goes against the teachings of the monks... the forefathers defined xintercept as "the value of y when x is 0" however, there is no y....

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0this is not a function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right... 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
 4 years ago
Best ResponseYou've already chosen the best response.0no. this is just not a function. nothing more; nothing less

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0it's a relation, but not a function

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0xintercepts occur in function only

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when did i say it is a function?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I said it was, my mistake

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0you were saying it had an xintercept

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0it has values for x...but no xintercept

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but the vertical line intercepts the xaxis...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes. LG you are right

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0http://mathworld.wolfram.com/xIntercept.html "The point at which a curve or FUNCTION crosses the xaxis (i.e., when in two dimensions)."

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0like i said. you got trolled

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so are you saying that a circle (which is a curve) does not have xintercepts?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0yes it doesnt. it's not a function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0circle is not a function

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0neither do ellipses

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0function = 1 to 1 value

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so why can't you call the where a vertical line crosses the the xaxis the xintercept?

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0any closed figure is not a function. it is a plane

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is going beyond the problem now... bottom line: x has two values

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i agree to agree... if that makes sense....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey LG... i thnk u scared off the asker.... or he/she got bored...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did she ask the problem to be graphed?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol jk. let us just leave it

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0i would like to quote myself "you got trolled" relations have xintercepts too http://en.wikibooks.org/wiki/Algebra/Intercepts

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes.... let's graph it....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this 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
 4 years ago
Best ResponseYou've already chosen the best response.0sorry man... i just miss talkin to LG....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u know u'd make a great ambassador for keeping the peace....:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so peace to everyone.....:)
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