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# integreation question quite hard watch

1. the gradient of a curve is given by dy/dx=8x-3x^2

given that the curve passes thru the origin find the equation of the tangent to the curve parallel to the x-axis?

2. the gradient of a curve is given by dy/dx=8x-3x^2

given that the curve passes thru the origin find the equation of the tangent to the curve parallel to the x-axis?

3. well firstly, if the tangent is parallel to the x-axis, what can we say about it's gradient, so what does that make dy/dx equal to?

Now when you integrate, you'd get a constant of integration. However knowing the curve passes through the origin, (0,0) we may be able to deal with that little bugger

Hope that helps, trying to point you the right way without telling you so if it's nonsense do ignore
4. Would it not be drawn at the minimum point of the function? - the tangent equation should be fairly simple as well because dy/dx = 0.
5. (Original post by ForGreatJustice)
well firstly, if the tangent is parallel to the x-axis, what can we say about it's gradient, so what does that make dy/dx equal to?

Now when you integrate, you'd get a constant of integration. However knowing the curve passes through the origin, (0,0) we may be able to deal with that little bugger

Hope that helps, trying to point you the right way without telling you so if it's nonsense do ignore
the gradient equals to 0 and when you intergrate you would get 4x^2-x^3 and stuck again now lol
6. y'=8x-3x^2
Integrating
y=4x^2-x^3+c
Substituting Pt (0,0) to find c
0=0-0+c
So
c=0
and y=4x^2-x^3

For a line to be parallel to the x axis y'=0

Thus
0=8x-3x^2
0=x(8-3x)
so x=0 and x=8/3

Substituting x=0 into y=4x^2-x^3
y=0

Substituting x=8/3 into y=4x^2-x^3
y=256/9 (might have miscalculated here)

m=0 so y-b=m(x-a) -> y-b=0

So y=256/9 or y=0

I'd be inclined to eliminate y=0 because it is the x axis so y=256/9

Check my calculations for mistakes 'cos I'm infamous for mistakes.

Hope this helps.

7. 8x-3x^2=0 >>> gets you and x value

the equation would be y=mx+c, but m=0 so would just be y=c
so from y=4x^2-x^3 you can find the y-value, which is the equation of the line
8. (Original post by R. Murray)
y'=8x-3x^2
Integrating
y=4x^2-x^3+c
Substituting Pt (0,0) to find c
0=0-0+c
So
c=0
and y=4x^2-x^3

For a line to be parallel to the x axis y'=0

Thus
0=8x-3x^2
0=x(8-3x)
so x=0 and x=8/3

Substituting x=0 into y=4x^2-x^3
y=0

Substituting x=8/3 into y=4x^2-x^3
y=256/9 (might have miscalculated here)

m=0 so y-b=m(x-a) -> y-b=0

So y=256/9 or y=0

I'd be inclined to eliminate y=0 because it is the x axis so y=256/9

Check my calculations for mistakes 'cos I'm infamous for mistakes.

Hope this helps.

FAIL
9. (Original post by DeanK2)
FAIL
?
10. (Original post by R. Murray)
?
I haven't checked it - I was just saying that posting up a full solution is frowned upon.

11. its inncorrect you get for the aNSWER y=256/27
12. (Original post by DeanK2)
I haven't checked it - I was just saying that posting up a full solution is frowned upon.

Oh... Maybe I'll edit out some numbers.

Silly me. Off I go with my editing hat on!

EDIT:
(Original post by sahil112)
its inncorrect you get for the aNSWER y=256/27
So it is - I've made a boo boo with my adding fractions. My calculator is upstairs and well, I can't be bothered...
13. i was trying this question, and have just thought
if 8x-3x^2 = 0
can you then say 8 - 3x = 0 (by dividing both sides by x)
because that rules out the possibility of x = 0
but seems 2 make sense
14. (Original post by crrrrrash)
i was trying this question, and have just thought
if 8x-3x^2 = 0
can you then say 8 - 3x = 0 (by dividing both sides by x)
because that rules out the possibility of x = 0
but seems 2 make sense
I don't get what you're trying to say. x = 0 and x = 8/3 are BOTH solutions of 8x - 3x^2 = 0. When you divide through by stuff, you can lose solutions, which is what you're doing when you divide through by x, rather than factorising it. It's like

x^2 + 5x + 6 = 0 can be solved in the following way
(x + 3)(x + 2) = 0
x = -3, -2

But we could just divide through by x + 3:

(x + 3)(x + 2) = 0
(x + 2) = 0
x = -2

Notice how we now lose the x = -3 solution. To solve 8x - 3x^2 = 0, you want to factorise the x

x(8 - 3x) = 0

And then note x = 0 and 8 - 3x = 0 are solutions.
15. (Original post by Swayum)
I don't get what you're trying to say. x = 0 and x = 8/3 are BOTH solutions of 8x - 3x^2 = 0. When you divide through by stuff, you can lose solutions, which is what you're doing when you divide through by x, rather than factorising it. It's like

x^2 + 5x + 6 = 0 can be solved in the following way
(x + 3)(x + 2) = 0
x = -3, -2

But we could just divide through by x + 3:

(x + 3)(x + 2) = 0
(x + 2) = 0
x = -2

Notice how we now lose the x = -3 solution. To solve 8x - 3x^2 = 0, you want to factorise the x

x(8 - 3x) = 0

And then note x = 0 and 8 - 3x = 0 are solutions.
haha
actually makes a lot of sense
silly me!

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