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April 16th, 2016, 07:35 AM  #1 
Newbie Joined: Apr 2016 From: Canada Posts: 2 Thanks: 0  Trying to figure out a standard form quadratic function from given info
Greetings; there is a quadratic relation question which I seek to engage with. The question is “determine the quadratic function that has given roots and passes through the given point. Express each function in standard form.” x=2±√5 (2,10)=(x,y) _____ Roots=c1 and c2 (+)=4.236067977 ()=0.236067977 Factored form Y=a(x+c1)(x+c2) 10=a (2+4.236067977) (2+(0.236067977)) 10=a(6.236067977)(1.76393202) 10=a10.99999998 10/10.99999998=a10.99999998/10.99999998 0.90909091=a ______ Standard form of the quadratic function 0=ax^2+bx+c a=0.90909091 , b=? , c=? From this point, I do not know which steps should be taken to solve for b and c . Last edited by skipjack; April 16th, 2016 at 08:21 AM. 
April 16th, 2016, 08:32 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,627 Thanks: 2077 
The quadratic equals a((x  2)²  5), which equals 5a when x = 2, so a = 2. Now rewrite 2((x  2)²  5) in standard form. 
April 16th, 2016, 09:43 AM  #3  
Newbie Joined: Apr 2016 From: Canada Posts: 2 Thanks: 0  Quote:
...For what reasons/why would would a quadratic/vertex form equation be a((x  2)²  5) does it have something to do with the square roots?... Also for what reasons/why would you say that 5a when x=2? Is it b/c?.. a((22)^25) a((0)^25) a(5) 5 ...? ...if I may guess, does it have something to do with...? x=2±√5 (2+√5/4.236067977,0) (2√5/0.236067977,0) (2,10)=(x,y) y=a(xh)^2+k ...? _ My apologies for coming off as discouraging while I am thankful for how what you did was encouraging. Last edited by skipjack; April 19th, 2016 at 01:10 AM.  
April 18th, 2016, 02:45 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 639 Thanks: 85 
(2, 10) is a given point on the curve. When x = 2, x  2 = 0, and it = 5a like skipjack said. Now 5a = 10, so a = 2.

April 19th, 2016, 01:08 AM  #5  
Global Moderator Joined: Dec 2006 Posts: 20,627 Thanks: 2077  Quote:
Using your original method (but with a sign error corrected), y = a(x  2 + √5)(x  2  √5) = a(x²  4x  1). Substituting (x, y) = (2, 10) gives 10 = a(5), so a = 2.  

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