What is the maximum value of 
such that the graph of the parabola 
has at most one point of intersection with the line 
+128475 Let y = (1/3)x^2 = g(x)
We are looking for the point where y = x + c is tangent to g(x)
Beause the slope of y = x + c = 1 we are looking for the point where the derivative of g(x) has a slope = 1.
So
g ' (x) = 2/3x ....set this = 1 and x = 3/2
And to find the y coordinate of this point we have y= (1/3)(3/2)^2 = (1/3)(9/4) = 9/12 = 3/4
So, since the point (3/2, 3/4) is also on y = x + c we have
3/4 = 3/2 + c subtract 3/2 from both sides
c = -3/4
Here's the graph.....https://www.desmos.com/calculator/ugbqgcgqrf
+128475 Let y = (1/3)x^2 = g(x)
We are looking for the point where y = x + c is tangent to g(x)
Beause the slope of y = x + c = 1 we are looking for the point where the derivative of g(x) has a slope = 1.
So
g ' (x) = 2/3x ....set this = 1 and x = 3/2
And to find the y coordinate of this point we have y= (1/3)(3/2)^2 = (1/3)(9/4) = 9/12 = 3/4
So, since the point (3/2, 3/4) is also on y = x + c we have
3/4 = 3/2 + c subtract 3/2 from both sides
c = -3/4
Here's the graph.....https://www.desmos.com/calculator/ugbqgcgqrf
