Question 131786
So far so good.  The vertex is at (3,2) and the axis of symmetry is {{{x=3}}}.  As to the maximum/minimum thing, all you need to do is determine whether the parabola is concave up (makes a valley), or concave down (makes a hill).  The sign on the lead coefficient tells you directly.  If the a in {{{ax^2+bx+c}}} is negative, the parabola is concave down and the vertex is a maximum, if it is positive, the parabola is concave up and the vertex is a minimum.


In this case, 4 is positive, so concave up and f(3) = 2 is a minimum.


As to sketching a graph, you already have one point, the vertex, so plot that.  Then you need to make a table of values for two or three x values on one side of the axis of symmetry.  Once you have done that you will have the same number of values for the other side of the axis because of the symmetrical property.


x = 4, f(4) = 4*16 - 24*4 + 38 = 102 - 96 = 6  => (4,6)
x = 5, f(5) = 4*25 - 24*5 + 38 = 138 - 120 = 18 => (5,18)
x = 6, f(6) = 4*36 - 24*6 + 38 = 182 - 144 = 38 => (6,38)

And, because of symmetry about {{{x=3}}}
x = 2, f(2) = f(4) = 6 => (2,6)
x = 1, f(1) = f(5) = 18 => (1,18)
x = 0, f(0) = f(6) = 38 => (0,38)


Plot your points and draw a smooth curve.  Done.


{{{drawing(800,800,-2,10,-2,40,
grid(1),
graph(800,800,-2,10,-2,40,4x^2 - 24x + 38),
locate(3.2,1.8,V(3,2)),
circle(3,2,.05),
circle(4,6,.05),
circle(5,18,.05),
circle(6,38,.05),
circle(2,6,.05),
circle(1,18,.05),
circle(0,38,.05)
)}}}