how do you write a quadratic function with zeros 3/7 and -4?

Question

3/7 and -4

Answer 1

  Postulate :Every quadratic of aX² + bX + c   has roots of m, n if and only if

                                              aX² +bX +c  can be  factored into  ( x -m ) ( X -n)

  accordingly :

f(x) = ( X +4) ( X -3/7) = X² + 4X -3/7X - 12/7

                                       7X² +28X - 3X - 12=

                                        7X² + 25X - 12

Answer 2

if they give you the zeroes (also called roots) you can get the function ax^2+bx+c

by using the product of the roots and the sum of the roots

 

the sum of the roots will come out to be -b/a

 

 

the product of the roots is c/a

 

 

so the sum of your roots in your case (3/7 - 4) would be -25/7

 

the product of your roots 3/7 * -4 would be -12/7

 

so we have sum of roots -25/7 = -b/a

                 product of roots -12/7 = c/a

 

you can match up the a, b and c

a=7  b=25 (watch your signs) and c=-12

7x^2+25x-12