tolong dijawab ya >.< ...... jika kurva parabola y = [tex] x^{2} [/tex] + bx + c melalui titik-titik (-3,1) dan (1,-2), maka [tex] \frac{c}{b} [/tex] =

[tex]y=x^2+bx+c \\ $Melalui (-3,1)$ \\ 1=9-3b+c \\ 3b-c=8 \\ \\ $Melalui (1,-2)$ \\ -2=1+b+c \\ b+c=-3 \\ \\ 3b-c=8 \\ b+c=-3 \\ -----$ +$ \\ 4b=5 \\ b=\frac{5}{4} \\ \\ b+c=-3 \\ \frac{5}{4}+c=-3 \\ c=-3-\frac{5}{4} \\ c=-\frac{17}{4} \\ \\ $Maka:$ \\ \frac{c}{b}=\frac{-^{17}/_4}{^5/_4} \\ \frac{c}{b}=-\frac{17}{5}[/tex]

(-3,1) =====> 1=[tex] (-3)^{2} +b(-3)+c=9-3b+c =====>-3b+c=-8 .......(1) [/tex]
(1,-2) =====> -2=[tex] 1^{2} +b(1)+c=1+b+c =====> b+c=-3 ..............(2)[/tex]
persmaan (1) dan (2)
-3b+c = -8
  b+ c= -3     
-4b    = -5
  b=5/4 maka c=-3-(5/4)=-17/4
[tex] \frac{c}{b} = \frac{ \frac{-17}{4} }{ \frac{5}{4} } = \frac{-17}{5} [/tex]