SOLUTION: For the quadratic function f(x)=3x^2+6x+2 a.)Graph the quadratic function by determining whether its graph opens up or down and by finding the vertex, axis of symmetry, y intercep

Question 1066249: For the quadratic function f(x)=3x^2+6x+2
a.)Graph the quadratic function by determining whether its graph opens up or down and by finding the vertex, axis of symmetry, y intercept, x intercepts, if any.
b.)The domain of f is
the range of f is
c.)determine where the function is increasing and where it is decreasing.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
f(x)=3x^2+6x+2
function opens upward because positive coefficient in front of x^2. It decreases from x=-oo to x=-1 and increases from x >-1.
domain is all x.
vertex x-value is -b/2a or -6/3(2)=-1
f(-1)=3-6+2=-1.
Vertex is (-1,-1)
axis of symmetry is x=-1
Range is [-1,oo)
y intercept is at f(0), and that = 2
x-intercepts need quadratic formula
x=(1/6)(-6 +/- sqrt (36-24))= (1/6)(-6 +/- 2 sqrt (3))= -1 +/- ((1/3) sqrt (3)). Numerically, they are -.42 and -1.58

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