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