Prabriti Tiwari Posted May 1, 2015 Report Share Posted May 1, 2015 The question goes like this, Find the coordinates of the point on the graph of y= x^2-x at which tangent is parallel to the line y= 5x. Can someone show me how it is done. Would help alot thanks. Reply Link to post Share on other sites More sharing options...
Emilia1320 Posted May 1, 2015 Report Share Posted May 1, 2015 First you differentiate it getting y' = 2x - 1 Then you plug 5 into equation in the place of y so it's 5 = 2x - 1 and solve for x giving you x = 3 Then plug 3 into original equation to obtain y-coordinate giving you y = 3^2 - 3 = 6 Coordinates are (3, 6) 3 Reply Link to post Share on other sites More sharing options...
ultimateone Posted May 1, 2015 Report Share Posted May 1, 2015 First you differentiate it getting y' = 2x - 1Then you plug 5 into equation in the place of y so it's 5 = 2x - 1 and solve for x giving you x = 3Then plug 3 into original equation to obtain y-coordinate giving you y = 3^2 - 3 = 6Coordinates are (3, 6)Great answer. But can someone please explain why in the second step, 5 was substituted and not 5x? Because the question says y= 5x, not y= 5. Sorry if this might be a stupid question haha - Maths isn't my thing. Reply Link to post Share on other sites More sharing options...
BrianHQ Posted May 1, 2015 Report Share Posted May 1, 2015 (edited) First you differentiate it getting y' = 2x - 1Then you plug 5 into equation in the place of y so it's 5 = 2x - 1 and solve for x giving you x = 3Then plug 3 into original equation to obtain y-coordinate giving you y = 3^2 - 3 = 6Coordinates are (3, 6)Great answer. But can someone please explain why in the second step, 5 was substituted and not 5x? Because the question says y= 5x, not y= 5. Sorry if this might be a stupid question haha - Maths isn't my thing. I think it's because you differentiate it to get the gradient y = 5x in y'= 5. Thus, she can equalize both terms. If y' = 5 and y' = 2x-1, then 5 = 2x -1.This only happens when we're talking about tangents as far as I'm concerned. I'm not that good at maths, but I believe this is why. Edited May 1, 2015 by BrianHQ Reply Link to post Share on other sites More sharing options...
Emilia1320 Posted May 1, 2015 Report Share Posted May 1, 2015 First you differentiate it getting y' = 2x - 1 Then you plug 5 into equation in the place of y so it's 5 = 2x - 1 and solve for x giving you x = 3 Then plug 3 into original equation to obtain y-coordinate giving you y = 3^2 - 3 = 6 Coordinates are (3, 6)It asked for a parallel line, so we take the gradient of y = 5x which is 5 and search for a point with same gradient on other curve. 1 Reply Link to post Share on other sites More sharing options...
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