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Give an example of a quadratic function that has zeroes at x=2 and x=4, and that takes the value 6 when x=3. Enter your answer in the expanded form "ax^2 + bx + c", where a,b,c are replaced by appropriate numbers.

 Oct 22, 2017
 #1
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P (x) = a (x - 2) (x - 4)     and we need to find "a"

 

And when x =  3  we have that 

 

 a( 3 - 2) (3 - 4)  = 6    simplify   and solve for "a"

 

a (-1) = 6  →   a =  -6

 

So we have

 

-6 [ x^2 - 6x + 8]  =

 

P(x) =  -6x^2 + 36x - 48

 

 

cool cool cool

 Oct 22, 2017

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