what quadractic function represents a 
parabola whose vertex is at (6,1) and 
passes through (4,9)

Any quadratic function can be expressed as

f(x) = a(x - h)² + k

where (h,k) is the vertex

So since (h,k) = (6,1),

f(x) = a(x - 6)² + 1

Since it passes through (4,9)

f(4) = 9

and also

f(4) = a(4 - 6)² + 1 
     = a(-2)² + 1 
     = a(4) + 1
     = 4a + 1

Since both 9 and 4a + 1
equal to f(4), we can
set them equal to each other:

9 = 4a + 1

Solve that and get a = 2

So substitute 2 for a in

f(x) = a(x - 6)² + 1

f(x) = 2(x - 6)² + 1

or if you like you can multiply
that out and simplify:

f(x) = 2(x - 6)(x - 6) + 1
     = 2(x² - 6x - 6x + 36) + 1
     = 2(x² - 12x + 36) + 1
     = 2x² - 24x + 72 + 1
     = 2x² - 24x + 73

Edwin