SOLUTION: I am given three points on a graph (14,0), (10.5,1), (6.5,3) and am asked to find the equation for the parabola that passes through these points. I understand that the parabola mu

The c value is not what you thought.  
You were thinking (14,0) was (0,14)
but it's just backwards from that.

Here we go from scratch:

A quadratic function has a parabola for its graph and
it must have an equation of thge form: 

y = ax² + bx + c

So plug each of the points in that equation.

Plug in (x,y) = (14,0)

y = ax² + bx + c

0 = a(14)² + b(14) + c

0 = 196a + 14b + c

-------

Plug in (x,y) = (10.5,1)

y = ax² + bx + c

1 = a(10.5)² + b(10.5) + c

1 = 110.25a + 10.5b + c

-------

Plug in (x,y) = (6.5,3)

y = ax² + bx + c

3 = a(6.5)² + b(6.5) + c

3 = 42.25a + 6.5b + c

-------

So we have this system

0 = 196a + 14b + c
1 = 110.25a + 10.5b + c
3 = 42.25a + 6.5b + c

or, putting the numbers on the right and lining up
the letters:

   196a +   14b + c = 0
110.25a + 10.5b + c = 1
 42.25a +  6.5b + c = 3

Can you solve that system?  If not post again.  

The answers to that system are a = , b = , c = 

So y = ax² + bx + c becomes

y = x² -x + 

It has vertex (, )

Its x-intercepts are (14, 0) and (41/2, 0)

Its y intercept is (0, 41/5)

Its axis of symmetry has the equation x = 

Its graph is this with the green line as the
axis of symmetry:



Edwin