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# Rep for the fastest reply- Differentiation watch

1. Here are the questions

1a. Find the coordinates O,A and B, the point of intersection of the curve with the x-axis. Hence, state the equations.

My attempt at finding the A intercept.
From my understanding if something intercepts with the x-axis, x=0

So subs back into original equation to get y=0.

So I know my 2 points are (0,0), differentiating to get the gradient I get this:

Putting x back in gives me the gradient which is 2, this is then put into the formula and I get the equation of the line as:

y=2x

My confusion is, how would I find the other points since x=0 y=0

2. Find the equation of the tangent to at the point where the curve meets the y-axis.

Any hints for this question

As always rep provided.
2. 1st.
1st.
Beat me to it. I hate it how tsr is seen as a marking service.
4. Substitute into the original thing. ie

(Original post by abstraction98)
Beat me to it. I hate it how tsr is seen as a marking service.
Help discussion, revision and homework help from KS3 and GCSE to degree level including for engineering, health-related subjects and sports science.
Thats one of the points of F38
5. (Original post by SimonM)
Substitute into the original thing. ie

subs 2x will give me x, I get it, but when I need to find just the coordinates of the letter, I get A as (0,0) which is right, then I don't understand how to get the coordinates of the other points.

1st.
That doesn't really help.
6. (Original post by Diaz)
subs 2x will give me x, I get it, but when I need to find just the coordinated of the letter, I get A as (0,0) which is right, then I don't understand how to get the coordinated of the other points.

What else can x be other than 0?

Edit

7. (Original post by Diaz)
That doesn't really help.
But it was the fastest reply
8. (Original post by Diaz)
Here are the questions

1a. Find the coordinates O,A and B, the point of intersection of the curve with the x-axis. Hence, state the equations.

My attempt at finding the A intercept.
From my understanding if something intercepts with the x-axis, x=0

So subs back into original equation to get y=0.

So I know my 2 points are (0,0), differentiating to get the gradient I get this:

Putting x back in gives me the gradient which is 2, this is then put into the formula and I get the equation of the line as:

y=2x

My confusion is, how would I find the other points since x=0 y=0

2. Find the equation of the tangent to at the point where the curve meets the y-axis.

Any hints for this question

As always rep provided.
I will help anyway. Factorise your initial equation. This requires finding the factors in a term with x^3 so you use special division i have forgotten the name of. That gives your x values (you should have been taught this at school)

By "get the equations", I assume you mean the equations of the tangents?

Once you ahve done this NOW differentiate as you have done correctly. Put x values in to get the gradient of each line. Shove the x values in to to coordinate form. All([xvalue], 0). Now we wheel out the formula:

y - b = m(x - a)

Lob the numbers from your intercepts in to get the tangents..

Hope that helped. Please be a bit more subtle in the future though
9. (Original post by SimonM)
Substitute into the original thing. ie

Thats one of the points of F38
Homework help - not giving the answers
10. (Original post by abstraction98)
Homework help - not giving the answers
I would say I am probably the most Nazi-ish person on the forum for not posting full solutions. Neither he or I asked/gave a full solution
11. Ban OP please for not giving rep for the fastest reply
12. I still have no idea what any of you are on about.

1. Why do I need to factorise? How to I obtain just the coordinates

2. Any hints?
13. (Original post by SimonM)
I would say I am probably the most Nazi-ish person on the forum for not posting full solutions. Neither he or I asked/gave a full solution
Stand corrected. The thread title implies it though
14. (Original post by Diaz)
I still have no idea what any of you are on about.

1. Why do I need to factorise? How to I obtain just the coordinates

2. Any hints?
Factorise you initial expression. The initial formula. This involves doing a special sort of division i have forgotten the name of. SImonM will know. You should have been taught it. This will tell you if the function divides nicely by (x + y) where y is a whole number (assuming this is simple).

For instance, if (x + 3) is a factor, an x intercept would be (-3, 0) (solving for (x + 3) = 0)
15. 1. It crosses the x axis when Y = 0. So, throw that in, and factorize the equation to find out when it does equal zero

2. It crosses the Y axis when x = 0, so if you differentiate it you can find the gradient of the tangent at that point....
16. (Original post by Diaz)
Here are the questions

1a. Find the coordinates O,A and B, the point of intersection of the curve with the x-axis. Hence, state the equations.

My attempt at finding the A intercept.
From my understanding if something intercepts with the x-axis, x=0

So subs back into original equation to get y=0.

So I know my 2 points are (0,0), differentiating to get the gradient I get this:

Shouldn't it be -6x and not -4x?
17. (Original post by LakHani_Hasnain)
Shouldn't it be -6x and not -4x?
I haven't been going through the numerics but agree entirely.
18. (Original post by Diaz)
I still have no idea what any of you are on about.

1. Why do I need to factorise? How to I obtain just the coordinates

2. Any hints?
You should be able to factorise x^3 equations.

Take out x as the common factor and then you get a quadratic...then solve. There is a simpler way, look on some of the websites...they'll be a little more insightful. Also i really cannot be bothered to write it all out for you.
19. I think the values of x are 3 and 0. If you got something else like a graph with the question I would think that would help alot. Post the whole question if possible.

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