Find parameter p so that y = 3x is a tangent line to the curve
y = x ² + p

Dear nitin,

• The slope of the tangent is equal to 3. At the point of tangency y ' = 3.
  
  y ' = 2 x = 3
  
  
• Solve the equation 2 x = 3 to find the x coordinate of the point of tangency.
  
  x = 3 / 2
  
  
• The point of tangency is on the line y = 3 x, therefore the y coordinate of the point of tangency is equal to
  
  y = 3 (3/2) = 9 / 2
  
  
• The point of tangency (3 / 2 , 9 / 2) is on the curve
  y = x ² + p and is therefore used to find p as follows
  
  9 / 2 = (3 / 2) ² + p
  
  
• Solve the above for p to obtain
  
  p = 9 / 4

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