Question

find a constant c such that the graph of x^2+11x+c in the xy plane has its vertex on the line y=x the hint said "complete the square"

Answer 1

Let's see... try adding 30 and subtracting 30, then getting (x+5)(x+6)+c-30... well, that's completing the square.

What I'd do is find y'=2x+11=0, x=-11/2, and then (-11/2)^2+11(-11/2)+c=-11/2. From here solving for c should be trivial.

Answer 2

x^2+11x+121/4-121/4+c
(x+11/2)^2+c-121/4
c-121/4=0
c=121/4

Answer 3

or
x^2+11x+c=x
x^2+10x+c=0
ax^2+bx+c=0
delta=b^2-4ac
delta=0 y=x line tangent of parabola
it is already said in the question
10^2-4*1*c=0
c=25