+0

# help

0
287
2
+1198

What is the maximum value of  such that the graph of the parabola $$y = \dfrac{1}{3}x^2$$ has at most one point of intersection with the line $$y = x+c?$$

Jul 30, 2019

#1
+8965
+4

Assuming the question is "What is the maximum value of  c  such that. . ."

$$y\ =\ \frac13x^2\\~\\ y\ =\ x+c$$

Equate both expressions of  y

$$\frac13x^2\ =\ x+c\\~\\ \frac13x^2-x-c\ =\ 0$$

This quadratic equation will have only one solution when the discriminant equals zero.

$$(-1)^2-4(\frac13)(-c)\ =\ 0\\~\\ 1+\frac43c\ =\ 0\\~\\ \frac43c\ =\ -1\\~\\ c\ =\ -\frac34$$

Jul 30, 2019

#1
+8965
+4

Assuming the question is "What is the maximum value of  c  such that. . ."

$$y\ =\ \frac13x^2\\~\\ y\ =\ x+c$$

Equate both expressions of  y

$$\frac13x^2\ =\ x+c\\~\\ \frac13x^2-x-c\ =\ 0$$

This quadratic equation will have only one solution when the discriminant equals zero.

$$(-1)^2-4(\frac13)(-c)\ =\ 0\\~\\ 1+\frac43c\ =\ 0\\~\\ \frac43c\ =\ -1\\~\\ c\ =\ -\frac34$$

hectictar Jul 30, 2019
#2
+1198
+1

thanks hactictar!

Jul 30, 2019