# Exercise Question from Maths book

1. Dec 3, 2015

### King_Silver

• Member warned about posting with no effort
1. The problem statement, all variables and given/known data
"Find all the points on the curve with equation 6y = 2x^3 + 3x^2 + 6x + 5 where the tangent is parallel to the line with equation 3x − y + 2 = 0

2. Relevant equations
No idea

3. The attempt at a solution
There are no exercises in the book even similar to this example which is fairly frustrating. Its under the differentiation chapter so it obviously involves differentiation. I can do that myself no bother but what is the method to find all points on the curve when a tangent is parallel to the line? Is there some sort of rule?

NOTE: This is not a graded assignment or even homework for that matter, it is revision for an exam I have in 5 days.

2. Dec 3, 2015

### micromass

Staff Emeritus
Can you find the tangent line to your curve in an arbitrary point $(a,b)$?

3. Dec 3, 2015

### King_Silver

Errrrrr.... I might be able to?
I need to find all points on the curve where the tangent is parallel to the line. I've tried finding a video on YouTube that works with a similar problem but I wasn't able to find anything for it.

4. Dec 3, 2015

### micromass

Staff Emeritus
How do you find the tangent line to the curve? Do you see the relation with derivatives?

5. Dec 3, 2015

### King_Silver

You differentiate the 6y - 2x^3 +3x^2 +6x +5
but I am not given the points on the curve I am asked to find them :/

6. Dec 3, 2015

### SteamKing

Staff Emeritus
Well, can you at least find the slope of the line 3x - y + 2 = 0?

7. Dec 3, 2015

### King_Silver

y'(x) = 3 is the slope of the line 3x -y +2 = 0 right? :)

8. Dec 3, 2015

### SteamKing

Staff Emeritus
Yes, this is correct.

Now, can you turn 6y = 2x^3 + 3x^2 + 6x + 5 into y = something ?

9. Dec 3, 2015

### King_Silver

Well if 6y = 2x^3 + 3x^2 +6x +5 then y = (2x^3 + 3x^2 +6x +5)/6 no?

10. Dec 3, 2015

### SteamKing

Staff Emeritus
That's correct.

Now do you understand that the derivative dy/dx for a function can be used to calculate the slope of a tangent to the function when given a particular value of x?

11. Dec 3, 2015

### King_Silver

Yea I understand that I just don't get how to find the points on the curve where the tangent is parallel to the line.

12. Dec 3, 2015

### SteamKing

Staff Emeritus
Well, have you calculated dy/dx for y = (2x^3 + 3x^2 +6x +5)/6 yet?

BTW: I appreciate the likes, but it's not necessary to like every reply.

13. Dec 3, 2015

### micromass

Staff Emeritus
How can you see whether two lines are parellel?

14. Dec 3, 2015

### King_Silver

If their slopes are identical? i.e. the exact same value yes?

15. Dec 3, 2015

### micromass

Staff Emeritus
Yes. So you need to find the points whose tangent line has slope is 3.

16. Dec 3, 2015

### King_Silver

dy/dx is x^2 + x + 1.

17. Dec 3, 2015

### SteamKing

Staff Emeritus
Now that you can find the slope along the function, for what values of x would dy/dx = 3?

18. Dec 3, 2015

### King_Silver

dy/dx would = 3 for....
x = 1
(1)^2 + 1 + 1 = 3. What does that prove in terms of the question? :)

19. Dec 3, 2015

### SteamKing

Staff Emeritus

Also, is x = 1 the only value which satisfies the equation dy/dx = 3 when dy/dx = x2 + x + 1 ?

20. Dec 3, 2015

### King_Silver

x = 1 and x = -2 (because (-2)^2 + (-2) + 1 => 4 -2 + 1 = 3)
so you have x = 1, x = -2.

Then you just do when x = 1, y = whatever you solve.
x = -2, y = whatever you solve
?