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Quadratic eq which takes the value (y) = 41 at x= -2 , the value (y) = 20 at x =5

when x = 2 -> a4-b2+c=y ...

how to find a,b,c,y

here is original question: Question's pic

Cool
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1 Answers1

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Your aren't supposed to find y. y is variable.

The quadratic equation is in the form $y = ax^2 + b^x + c$. When x = -2 you have y = 41 so $41 = a*(-2)^2 +b*(-2) + c$. When x = 5, y= 20 so $20 = a*5^2 + b*5 + c$. When x = 2, y = $a4-b2+c$ so $a4-b2+c = a*2^2 + b*2 + c$.

You have three equations and three unknowns, a, b, c. So you should be able to solve a, b, c and express $y = a*x^2 + b*x +c$ with the proper values of a, b, c.

fleablood
  • 124,253
  • i thought, it'll be solved by system of equations like

    "4 -2 1 41 25 5 1 20"

    i tried, but variable' < equations and i couldnt find proper values of variables...

     – Cool Oct 09 '15 at 06:35