Question 824771
Why does the "c" parameter in standard form, f(x) = ax^2+bx+c, not equal the "k" parameter in vertex form, y = a(x-h)^2+k?  
<pre>
        f(x) = ax˛+bx+c
           y = a(x-h)˛+k

Set f(x) identically equal to y

Then for every value of x,

    ax˛+bx+c = a(x-h)˛+k   

In particular it is true when x=0, so

a(0)˛+b(0)+c = a(0-h)˛+k
           c = ah˛+k

So c is not equal to k except when a or h is 0.
If h=0 then the vertex is (0,k) which is a point
on the y-axis and if a=0 the equation is not a 
quadratic, but only f(x) = c and y = k, a
horizontal line. 

Edwin</pre>