Find a formula for a call dragon function with 0 at X =3 and x=-5

quadratic function
zeros: x = -5 and 3
y-intercept at (0,12)
Since, the factors are:
(x - (-5) )
(x - 3)
We write the function as it follows
[tex]f(x)=a(x+5)(x-3)[/tex]we can find a from a given point, for example (0,12)
[tex]\begin{gathered} f(0)=a\cdot(0+5)(0-3)=12 \\ a\cdot5\cdot(-3)=12 \\ a=-0.8 \end{gathered}[/tex]therefore, the function is
[tex]f(x)=-0.8(x+5)(x-3)[/tex]we can rewrite the expression as it follows:
[tex]\begin{gathered} -0.8\mleft(x+5\mright)\mleft(x-3\mright) \\ =-0.8\mleft(x^2+5x-3x-15\mright) \\ =-0.8\mleft(x^2+2x-15\mright) \\ =-0.8x^2-0.8\cdot\: 2x-0.8\mleft(-15\mright) \end{gathered}[/tex]simplify
[tex]=-0.8x^2-1.6x+12[/tex]