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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. Tags figure, form, function, info, quadratic, standard Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post abc98 Algebra 10 September 20th, 2015 05:49 PM PerfectTangent Algebra 2 February 25th, 2014 09:18 PM amirus14 Algebra 3 January 23rd, 2012 01:54 PM falmarie Algebra 1 February 10th, 2011 10:35 AM Wojciech_B Linear Algebra 0 December 7th, 2009 01:52 PM

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