SOLUTION: A quadratic function has an axis of symmetry of x=4, a y-intercept of 33 and also passes through the point, P(5,3). Write the equation of this function in the form of y=ax^2+bx+c.

Let's draw what we have,  Since its y-intercept is 33, it passes
through the point (0,33). We will draw its axis of symmetry and plot
the points (0,33) and (5,3).




Since the line x=4 is the axis of symmetry and since the point
(5,3) is on the graph, its reflection across the line x=4 is
also a point on the graph.  That point is (3,3).  Also the
reflection of the y-intercept (0,33) across the line of symmetry 
is also a point on the graph.  That point is (8,33).  So we have 
four points on the parabola and we only need three.

   


So we pick any three of the four points and substitute them in



Substituting in (0,33)




So we already have c=33:



Substituting (8,33) in that






Substituting (3,3) in that








Solve that system and get a=2 and b=-16

So the parabola's equation is



and here is its graph: 



Edwin