for y=ax²+bx+c
if a>0 then it opens up
if a<0 then it opens down
hmm, y intercept is where x=0
x intercept is where y=0
so
f(x)=3x²+4x+16
3>0 so opens up
y intercep
x=0
3(0)²+4(0)+16
16
y intercept is y=16
x intercept
0=f(x)=3x²+4x+16
there are no real solutions since the discriminant is negative
no x intercepts or no zeroes
so
opens up
y intercept of 16
no zeroes