what is the equation of a quadratic in standard form, given the zeros x=2 and x=-4 and the point (1,-5).
A) F(x)=X^2-2x-8
B)F(x)=X^2+6x+8
C)F(x)=X^2+2x-8
D)F(x)+X^2-6x+8
whenever I do this I always get a -X^2 and obviously that isn't a choice so it's wrong.

Question

what is the equation of a quadratic in standard form, given the zeros x=2 and x=-4 and the point (1,-5).
A) F(x)=X^2-2x-8
B)F(x)=X^2+6x+8
C)F(x)=X^2+2x-8
D)F(x)+X^2-6x+8
whenever I do this I always get a  -X^2 and obviously that isn't a choice so it's wrong.

anonymous · Answer

@21664ryker

anonymous · Answer

equation of the function with zeroes x=2 and x=-4 are given by (x-2)(x+4)=0 this implies x^2-2x+4x-8=0 this implies x^2+2x-8=0. Hence the required form is F(x)=x^2+2x-8 option C is correct. Plug in x=1, and you see here F(1)=-5 and hence the solution is perfect