y-intercept occurs at x=0, and x-intercept occurs at f(x) =0
f(x) =3x^2 -20x+12
f(0) = 12
(5) (0,12)
3x^2 -20x+12=0
D = b^2 - 4ac
D = (-20)^2 - 4*3*12 = 256
x = (-b +- sqrt(D))/2a
x = (20 +- 16)/2*3
so x = 2/3 or x= 6
(6) (2/3,0)(6,0)
f(x) = 15+2x-x^2
f(0) =15
7) (0,15)
15+2x-x^2=0
using quadratic formula we get:
x= -3 or x=5
(8) (-3,0)(5,0)