What is the maximum value of c such that the graph of the parabola y = (1/3)x^2 has at most one point of intersection with the line y = x+c?

Answer 1

x+c=1/3 x^2

x^2-3x+3c=0
making it one solution,
x^2-3x+(3/2)^2=0
or 3c=(9/4) or 3/4
x-3x-9/4=0
(x-3/2)^2=0 single solution occurs at that x=3/2 and max c is 3/4

check my work and thinking.

Answer 2

That's what I tried, but it's wrong. Idk why.

Answer 3

It is actually -3/4