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C1 Differentiation

Hi, how do I do question 8iii?

http://i.imgur.com/iomYmdR.png

I differentiated the equation to 9x^2 - 2x^-2 - 7 = 0

Then I used the logic that x = 0 and y = -7, yet the mark scheme said that the answer was y = -2.

Thank you.
(edited 8 years ago)
Original post by Peppercrunch
Hi, how do I do question 8iii?

I differentiated the equation to 9x^2 - 2x^-2 - 7 = 0

Then I used the logic that x = 0 and y = -7, yet the mark scheme said that the answer was y = -2.

Thank you.


We can't see the question, can you post an image/link?
Question:
Original post by Peppercrunch
Question:


You cannot substitute x = 0 into the differential to find the y value, it is true that x = 0 at Q but you need to find the equation of the tangent.
Original post by lizard54142
You cannot substitute x = 0 into the differential to find the y value, it is true that x = 0 at Q but you need to find the equation of the tangent.


How would I find the equation of the tangent?
Reply 5
Avatar for aoxa
aoxa
11
Original post by Peppercrunch
How would I find the equation of the tangent?


Sub x=1 into the original equation of the curve. As the tangent at the stationary point will have a gradient of 0, it will just be a line, and so the value when the tangent crosses the x axis will be whatever you get when you sub x=1 into the original equation of the curve.
Original post by aoxa
Sub x=1 into the original equation of the curve. As the tangent at the stationary point will have a gradient of 0, it will just be a line, and so the value when the tangent crosses the x axis will be whatever you get when you sub x=1 into the original equation of the curve.


Oh, thank you. Didn't think it would be that easy.
Reply 7
Avatar for aoxa
aoxa
11
Original post by Peppercrunch
Oh, thank you. Didn't think it would be that easy.


You're welcome - if you're not sure how hard it should be, just look at the number of marks, if you're doing loads of working for a 1/2 mark question, you're probably missing something obvious :smile:

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