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finding the quadratic formula

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1 Answer

(1.7)? It doesn't make sense here.

Actually, three points, such as (2,14) (3,28) (4,56),  can determine a unique quadratic equation.

If you are allowed to use calculator, then use "STAT" to do curve fitting. The answer is

f(x) = 7x^2 - 21x + 28

If you are not allowed to use calculator, then let f(x) = ax^2 + bx + c

Plug in all three given points,

14 = 4a+2b+c ......(1)

28 = 9a+3b+c ......(2)

56 = 16a+4b+c ......(3)

(1)+(3)-2*(2): 14 = 2a => a = 7

(2)-(1): 14 = 5a+b => b = 14-5a = -21

c = 14-4a-2b = 14-28+42 = 28

Comments

Hi Robert. I believe, there is something wrong with last y-coordinate, or quadratic equation is not exist for all 4 points:
(2) - (1): 14 = 5a + b
(3) - (2): 28 = 5a + b

3) - (2): 28 = 5a + b

It should be 3) - (2): 28 = 7a + b

Thank you, Robert. Why did I even doubt you work? Shame on me! :( 

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