Find an equation for the quadratic that passes through the points (−1,8), (2,−4), and (−6,−12).
+208 y=ax^2+bx+c
first substitute (-1, 8) (1 for x and 8 for y)
8=a-b+c
then substitute (2, -4) (2 for x and -4 for y)
-4=4a+2b+c
then substitute (-6, -12) (-6 for x and -12 for y)
-12=36a-6b+c
now we have these three equations:
8=a-b+c
-4=4a+2b+c
-12=36x-6x+c
so a=-1
b=-3
and c=6
and the quadratic equation's equation is y=-x^2-3x+6
JP
