My Math Forum Trying to figure out a standard form quadratic function from given info

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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: 20,973 Thanks: 2224 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
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Quote:
 Originally Posted by skipjack 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.
Hello skipjack; while I do appreciate how you responded, I don't understand how you arrived at the equation which you wrote of in the response...

...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((2-2)^2-5)

a((0)^2-5)

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(x-h)^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: 664 Thanks: 87 (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
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Quote:
 Originally Posted by 1relaxing For what reasons . . . does it have to do with the square roots?
Yes, x = 2 ± √5 implies (x - 2)² = 5, so a((x - 2)² - 5) = 0.

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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