Question 949638
we are dealing with parabolas
1) red curve is a, green curve is b, and blue curve is c
{{{ graph( 300, 200, -10, 10, -10, 10, 4x^2-12x, 4-2x-x^2, x^2-3.14) }}}
2) we are given the function f(x) = 3x^2 -5x +2
the graph of this function looks like
{{{ graph( 300, 200, -10, 10, -10, 10, 3x^2 -5x +2) }}}
since the x^2 term is positive, this parabola opens upward, therefore the x value for the axis of symmetry will give us the least value
axis of symmetry is given by
x = -b / (2*a) = -(-5) / (2*3) = 5/6
now substitute for x in our function
f(5/6) = 3*(5/6)^2 -5*(5/6) +2 = -1/12
therefore the value for x is 5/6 and the least value of f(x) is -1/12
range of this function, f(x) = 3x^2 -5x +2, is f(x) > or = -1/12
3) We are given
y=a(x-p)^2+q. Its turning point is located at (1,4). the graph also contains point (1,3).  This is a parabola that curves upward and the turning point tells us
i) x = -b/(2a) = 1 and
4 = a(1-p)^2 +q
the additional point tells us that
3 = a(1-p)^2 +q
The definition of a function is
"A function is a special relationship where each input has a single output", therefore this can not be a function since (1,4) and (1,3) are both on the graph of the function.  Also (1,4) can not be the turning point since (1,3) is on the graph and f(x) = 3 is < f(x) = 4