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?
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