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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

Best Answer 

 #1
avatar+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.

 

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

 

Check: https://www.desmos.com/calculator/0rnbjixeej

 Jul 30, 2019
 #1
avatar+8965 
+4
Best Answer

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.

 

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

 

Check: https://www.desmos.com/calculator/0rnbjixeej

hectictar Jul 30, 2019
 #2
avatar+1198 
+1

thanks hactictar!

 Jul 30, 2019

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