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tangents and differentiation c1

The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by gomc
The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?


assuming your working are correct put x=1 into the equation of the curve to get the corresponding y coodinate
Reply 2
Avatar for Zacken
Zacken
22
Original post by gomc
The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?


So you have x=1x=1, what value does your curve take when you sub x=1x=1 into it?

Your final answer will be P(1, something )P(1, \, \mathrm{ something }\, )
Reply 3
Avatar for Zacken
Zacken
22
Original post by TeeEm
assuming your working are correct put x=1 into the equation of the curve to get the corresponding y coodinate


Beaten again. :tongue:
Original post by gomc
The tangent at point P on the curve y= (3x-5)(x+2) is parallel to the line y = 7x+1. Find the co-ordinates of point P.

I've differentiated the first equation to get 6x+1 and made this equal to 7, then solved to find x, which is 1. Now what?



The gradient of the tangent to the curve at point P is equal to the gradient of the line y = 7x + 1.

You need to put the differential of the first equation equal to the gradient of the line and solve for x.
Does it make sense why you would do that?
Never mind, I can't read, apparently. Ignore me! :P
(edited 8 years ago)
Reply 5
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gomc
OP
11
but when i plot the two graphs they dont cross at (1,-6)?? no where near
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by Zacken
Beaten again. :tongue:


is all this LaTeX typing ...
Reply 7
Avatar for Zacken
Zacken
22
Original post by gomc
but when i plot the two graphs they dont cross at (1,-6)?? no where near


Because the line y=7x+1y = 7x + 1 is parallel to the tangent at PP, it's not the actual tangent and hence will not cross the line.

The tangent to the curve at PP is given by y+6=7(x−1)  ⟺  y=7x−13≠7x+1y + 6 = 7(x-1) \iff y = 7x - 13 \neq 7x+1.
(edited 8 years ago)
Reply 8
Avatar for gomc
gomc
OP
11
Original post by Zacken
Because the line y=7x+1y = 7x + 1 is parallel to the tangent at PP, it's not the actual tangent and hence will not cross the line.

The tangent to the curve at PP is given by y+6=7(x−1)  ⟺  y=7x−13≠7x+1y + 6 = 7(x-1) \iff y = 7x - 13 \neq 7x+1.

god how did i miss that im so tired lol thanks though
Reply 9
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Zacken
22
Original post by gomc
god how did i miss that im so tired lol thanks though


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