You can put this solution on YOUR website! what is the intercepts, axis of symmetry and vertex of y=3x^2+12x+16
-----------
y-intercept: Let x = 0, then y = 16
------------
x-intercepts:
Solve 3x^2+12x+16 = 0
x = [-12 +- sqrt(144 - 4*3*16)]/6
---
x = [-12 +- sqrt(-48)]/6
Since the discriminant is negative there are no x-intercepts.
------------
Vertex occurs where x = -b/2a = -12/(2*3) = -2
f(-2) = 3(4)+12(-2)+16 = 28-24 = 4
---
Vertex (-2,4)
---
Axis of symmetry is x=-b/2a = -2
--------------------------------------
=========================================
Cheers,
Stan H.
(