Describe the direction of the parabola and determine the y-intercept and zeros. f(x) = 3x^2 + 4x +16

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

D