Question 355341
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. 
<pre>
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).

{{{drawing(11000/41,1000,-2,9,-1,40,

line(0+.1,33,0-.1,33), line(0,33+.1,0,33-.1),line(0+.1,33-.1,0-.1,33+.1),line(0+.1,33+.1,0-.1,33-.1), green(line(4,-10,4,50)),

line(5+.1,3,5-.1,3), line(5,3+.1,5,3-.1),line(5+.1,3-.1,5-.1,3+.1),line(5+.1,3+.1,5-.1,3-.1), locate(5,3, "(5,3)"),


graph(11000/41,1000,-2,9,-1,40), 

locate(0,33, "(0,33)")



 )}}}


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.

{{{drawing(11000/41,1000,-2,9,-1,40,

line(0+.1,33,0-.1,33), line(0,33+.1,0,33-.1),line(0+.1,33-.1,0-.1,33+.1),line(0+.1,33+.1,0-.1,33-.1), green(line(4,-10,4,50)),

line(5+.1,3,5-.1,3), line(5,3+.1,5,3-.1),line(5+.1,3-.1,5-.1,3+.1),line(5+.1,3+.1,5-.1,3-.1), locate(5,3, "(5,3)"),

line(3+.1,3,3-.1,3), line(3,3+.1,3,3-.1),line(3+.1,3-.1,3-.1,3+.1),line(3+.1,3+.1,3-.1,3-.1), locate(3,3, "(3,3)"),

line(8+.1,33,8-.1,33), line(8,33+.1,8,33-.1),line(8+.1,33-.1,8-.1,33+.1),line(8+.1,33+.1,8-.1,33-.1), locate(6,33, "(8,33)"),

graph(11000/41,1000,-2,9,-1,40), 

locate(0,33, "(0,33)")



 )}}}   


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

{{{y=ax^2+bx+c}}}

Substituting in (0,33)

{{{33=a(0)^2+b(0)+c}}}
{{{33=c}}}

So we already have c=33:

{{{y=ax^2+bx+33}}}

Substituting (8,33) in that

{{{33=a(8)^2+b(8)+33}}}
{{{33=64a+8b+33}}}
{{{0=64a+8b}}}
{{{0=8a+b}}}

Substituting (3,3) in that

{{{3=a(3)^2+b(3)+33}}}
{{{3=9a+3b+33}}}
{{{-30=9a+3b}}}
{{{-10=3a+b}}}

{{{system(0=8a+b,-10=3a+b)}}}

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

So the parabola's equation is

{{{y=2x^2-16x+33}}}

and here is its graph: 

{{{drawing(11000/41,1000,-2,9,-1,40,

line(0+.1,33,0-.1,33), line(0,33+.1,0,33-.1),line(0+.1,33-.1,0-.1,33+.1),line(0+.1,33+.1,0-.1,33-.1), green(line(4,-10,4,50)),

line(5+.1,3,5-.1,3), line(5,3+.1,5,3-.1),line(5+.1,3-.1,5-.1,3+.1),line(5+.1,3+.1,5-.1,3-.1), locate(5,3, "(5,3)"),

line(3+.1,3,3-.1,3), line(3,3+.1,3,3-.1),line(3+.1,3-.1,3-.1,3+.1),line(3+.1,3+.1,3-.1,3-.1), locate(3,3, "(3,3)"),

line(8+.1,33,8-.1,33), line(8,33+.1,8,33-.1),line(8+.1,33-.1,8-.1,33+.1),line(8+.1,33+.1,8-.1,33-.1), locate(6,33, "(8,33)"),

graph(11000/41,1000,-2,9,-1,40,2x^2-16x+33), 

locate(0,33, "(0,33)")



 )}}}

Edwin</pre>