Let c be a real number. What is the maximum value of c such that the graph of the parabola \(y=\frac{1}{3}x^2\) has at most one point of intersection with the line \(y=x+c\)?

dolphinia May 8, 2023

#1**0 **

The parabola and the line will intersect at most once if the line does not pass through the vertex of the parabola. The vertex of the parabola is at (0,0). The line will pass through the vertex if c=0. Therefore, the maximum value of c such that the graph of the parabola y=x^2/3 has at most one point of intersection with the line y=x+c is 0.

Guest May 8, 2023

#3**0 **

What I found was **c = –0.75**

I don't know how to solve this algebraically, so

I went to Desmos and graphed both curves, and

Desmos let me add a slider to vary the value of c.

The line is tangent to the parabola about (1**.**33, 0**.**60)

which I estimated by eyeballing it.

.

Guest May 9, 2023

#4**+1 **

The line will touch the parabola at just one point where the line is tangent to the parabola, at which point their slopes are equal.

Slope of straight line = 1

Slope of parabola = 2x/3

So the x-coordinate is given by 2x/3 = 1 or x = 3/2

At this point the value of y is y = (1/3)*(3/2)^2 or y = 3/4

So, using the straight line equation: 3/4 = 3/2 + c or c = -3/4

Alan May 9, 2023