Question

A.) f(x)=3x^2-18x+27.

Answer 1

A)
Vertex:(3,0)
x-intercepts: (3,0) 
y-intercepts: (0,27)

B)
Vertex:(0,-9)
x-intercepts:none 
y-intercepts: (0,-9)

Answer 2

(b) is relatively easy.  Since there is no middle term, the graph is that of a parabola with vertex at (0, -9), opening downward due to the (-) sign

In (a), we have 3x^2-18x+27 = f(x).  3 can be factored out:  f(x) =3(x^2-6x+9).

Can you see that this is already a perfect square?  So f(x) = 3(x-3)^2 + 0.

The vertex is at (3, 0).  Place a black dot at that point on graph paper.  Because the leading coeff. of 3x^2-18x+27 is +, the parabola opens UP.  The y-intercept is found by letting x=0:  f(0) = 3(0)^2 - 18(0) + 27 => (0,27).