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# Showing line is a tangent of a curve watch

1. Show that the line 3x+y -2 = 0 is a tangent to the curve y=(4x-3)(x-2) and find the point of contact.

I just need to find the point of intersection right, so

======

I first rearrange the line to give :
y= 2/3x .

I then make it equal the curve, when expanded is 4x^2-11x+6 .

2/3x = 4x^2-11x+6

But I'm confused about bringing over the denominator of 3x to the other side, and I end up with a wrong answer.
2. I'm not sure how you got from but it's certainly not right!

You need to show that they intersect and that the gradient of the line is equal to the gradient of the tangent at the point of intersection.
3. (Original post by Gelato)
Show that the line 3x+y -2 = 0 is a tangent to the curve y=(4x-3)(x-2) and find the point of contact.

I just need to find the point of intersection right, so

======

I first rearrange the line to give :
y= 2/3x .

I then make it equal the curve, when expanded is 4x^2-11x+6 .

2/3x = 4x^2-11x+6

But I'm confused about bringing over the denominator of 3x to the other side, and I end up with a wrong answer.
3x+y -2 = 0 does not reaarange to y= 2/3x
4. sorry but how did you get 2/3x ? Isn't it just -3x+2=y or have i missed something?
5. oh this is how i done it :

3x+y - 2 = 0

I bring -2 to right to give 3x+y = 2

Then to get Y , bring 3x over to the right ?
6. (Original post by Gelato)
oh this is how i done it :

3x+y - 2 = 0

I bring -2 to right to give 3x+y = 2

Then to get Y , bring 3x over to the right ?
which makes y = 2 - 3x ....
7. Oh darn it , my handwriting made it look like a division , as I took a space under it

Thanks I got it now.
8. (Original post by Gelato)
oh this is how i done it :

3x+y - 2 = 0

I bring -2 to right to give 3x+y = 2

Then to get Y , bring 3x over to the right ?

So it would be y= 2-3x

Not dividing by 3x You'd be right it was multiplying..
9. Thanks for the help . Solved

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