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Can anyone help me with this differentiation problem!?!

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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.

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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)

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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)

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.

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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)

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 by BrianHQ

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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.
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