Answer:
y=x²+x-12
Step-by-step explanation:
The quadratic standard form is given by this formula: ax² +bx +c =0.
However all we have is the zeros of this function.
1) So let's start by plugging into x the zeros. We will have two linear functions
[tex]y=ax^{2} +bx+c\\\\ y=a(3)^{2} +3b+c\\ y=9a+3b+c \\ y=16a+4b+c[/tex]
2) Solving the system of linear equations, by the Addition Method:
[tex]\left \{{{9a+3b+c=0} \atop {16a-4b+c=0}} \right.\\ \left \{{{9a+3b+c*(-1)=0} \atop {16a-4b+c=0}} \right\\ \left \{{{-9a-3b-c=0} \atop {16a-4b+c=0}} \right. \\ \\ 7a-7b=0\\ a=b\\ \\9+3+c=0\\ c=-12[/tex]
Therefore a=b and c=-12 and x=-4 and x=3
Since a=b, let's assume a=1 and b=1 and as c=-12 is the product of -4*3 hence we can rewrite this equation grouping it, and then test it:
[tex]y=(x+4)(x-3)\\ y=x^{2} -3x+4x-12\\ y=x^{2} +x-12[/tex] Standard form
