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April 16th, 2016, 07:35 AM   #1
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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.
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April 16th, 2016, 08:32 AM   #2
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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.
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April 16th, 2016, 09:43 AM   #3
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Quote:
Originally Posted by skipjack View Post
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.
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April 18th, 2016, 02:45 PM   #4
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(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.
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April 19th, 2016, 01:08 AM   #5
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Quote:
Originally Posted by 1relaxing View Post
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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