Question 626766
 determine the quadratic function of f whose vertex is (4,3) and passes through
(3,1) put in f(x)=ax^2+bx+c form
"Vertex form" of a quadratic:
y = a(x-h)^2 + k 
the problem gives us everything except the 'a'.  Plug in given info and solve for 'a':
y = a(x-h)^2 + k 
1 = a(3-4)^2 + 3
1 = a(-1)^2 + 3
1 = a(1) + 3
1 = a + 3
-2 = a
.
so now we have:
y = -2(x-4)^2 + 3
.
Now, we reorganize to the right form:
y = -2(x-4)^2 + 3
y = -2(x-4)(x-4) + 3
y = -2(x^2-8x+16) + 3
y = -2x^2+16x-32 + 3
y = -2x^2+16x-29 (this is the form they want)