# a level "equations of lines" question

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#1
hi guys, Im stuck on part b of a question and would like some help please. it is:
A curve has equation y=x^2(4x+5) and the point P has coordinates (-1,1)
(a) Prove that the tangent to the curve at P has equation y=2x+3 (ive done this part )
(b) Show that this tangent intersects the curve again at the point with x-coordinate equal to 3/4.

i think you make the two equations equal to start off but idk whats next.
0
2 years ago
#2
Well once you've equated the two functions, you've effectively just made a new equation that you can solve.

, right?

You've got a cubic equation there. The roots of that equation will tell you the x-values for the points of intersection of the two functions.
The reason you can solve it is because you've been told one root - so you'll have to do polynomial long division.
1
#3
thanks!
(Original post by Sinnoh)
Well once you've equated the two functions, you've effectively just made a new equation that you can solve.

, right?

You've got a cubic equation there. The roots of that equation will tell you the x-values for the points of intersection of the two functions.
The reason you can solve it is because you've been told one root - so you'll have to do polynomial long division.
0
2 years ago
#4
Not sure if you've written the function correctly.

From the given equation of the curve, when x=-1. y=-9, which will give tangent as y=22x+13

And, if x=1 (maybe you got your coords wrong way round?), then y = -1
The tangent to the curve at this point is y = 2x - 3, but this tangent meets the curve again....
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#5
im not too sure what you mean but the question ive written is straight from the textbook and the only mistake I can see is that the other person who gave an answer is slightly wrong because they wrote 8x^3 when its actually 4x^3

oh and ive got the answer now lol
(Original post by begbie68)
Not sure if you've written the function correctly.

From the given equation of the curve, when x=-1. y=-9, which will give tangent as y=22x+13

And, if x=1 (maybe you got your coords wrong way round?), then y = -1
The tangent to the curve at this point is y = 2x - 3, but this tangent meets the curve again....
0
2 years ago
#6
i've graphed the equation that you gave, and included the tangents that I mentioned in earlier post.

Based on this, I'd be interested to see your result & how on earth did you manage to get y = 2x+3 as the eqn of the tangent?
im not too sure what you mean but the question ive written is straight from the textbook and the only mistake I can see is that the other person who gave an answer is slightly wrong because they wrote 8x^3 when its actually 4x^3

oh and ive got the answer now lol
0
#7
I expanded y, differentiated it and substituted -1 to find the gradient at that point and got 2, used the equation y-y1=m(x-x1) to find the equation of the tangent and simplified it
(Original post by begbie68)
i've graphed the equation that you gave, and included the tangents that I mentioned in earlier post.

Based on this, I'd be interested to see your result & how on earth did you manage to get y = 2x+3 as the eqn of the tangent?
0
2 years ago
#8
ok. so you've now altered your original eqn. but you've chosen not to admit to that.
0
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