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    Okay. So I have these 4 questions and have no idea why my answers are wrong, howver I will only show you the first one as it take forever to write this out & I have used the same principle on all of them:
    find the equation of the tangents to the curves at given points:
    2x^2 +9 @ x=2
    for this I did dy/dx = 2*2 x^2-1
    so m = 4x
    then I went back and used x=2 into the raw form
    y=2x^2+9
    2*(2^2) + 9
    y= 8+9
    y=17
    therefore
    17 = 4x + c
    x=2
    so
    17 = 4*2 +c
    17 = 8 + c
    c=9
    equation of tangent = y = 4x+9
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    You have to use y-y1 = m(x-x1) to get your equation
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    (Original post by Danny.L)
    Okay. So I have these 4 questions and have no idea why my answers are wrong, howver I will only show you the first one as it take forever to write this out & I have used the same principle on all of them:
    find the equation of the tangents to the curves at given points:
    2x^2 +9 @ x=2
    for this I did dy/dx = 2*2 x^2-1
    so m = 4x

    then I went back and used x=2 into the raw form
    y=2x^2+9
    2*(2^2) + 9
    y= 8+9
    y=17
    therefore
    17 = 4x + c
    x=2
    so
    17 = 4*2 +c
    17 = 8 + c
    c=9
    equation of tangent = y = 4x+9
    You forgot to find the value of the differential at x = 2.
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    (Original post by Jordan97)
    You have to use y-y1 = m(x-x1) to get your equation
    Both methods can be used. The value of m is incorrect.
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    (Original post by morgan8002)
    You forgot to find the value of the differential at x = 2.
    Okay, so could someone quote me exactly where I made my mistake. It's late and I've done 25 pages of notes on differentiation, so I'm having a hard time comprehending the maths terminology as I'm so worn out.
    Thanks
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    (Original post by Danny.L)
    Okay, so could someone quote me exactly where I made my mistake. It's late and I've done 25 pages of notes on differentiation, so I'm having a hard time comprehending the maths terminology as I'm so worn out.
    Thanks
    I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.
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    (Original post by morgan8002)
    I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.
    17=4x+c
    x=2
    4x = 4*2
    17 = 8 + c
    17-8 = c
    c=9
    I dont understand what you mean. I thought that is putting the x value into the differentiated gradient
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    (Original post by morgan8002)
    I bold coded in my previous reply to make it clear. You didn't substitute x = 2 into the differential to get the gradient.
    after doing the nx^nx-1 that gives you the gradient of any point on that line. It is 4x
    so I take that down and use it in the y=mx+c
    17 = 4x +c
    I cannot see what is wrong with this and dont understand what to do. I am seemingly getting some right and some wrong with the above method, it isnt wrong 100% of the time but neither is it right.
    I am yet to be taught this, I'm learning on my own, so I could use some guidence
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    (Original post by Danny.L)
    17=4x+c
    x=2
    4x = 4*2
    17 = 8 + c
    17-8 = c
    c=9
    I dont understand what you mean. I thought that is putting the x value into the differentiated gradient
    \frac{dy}{dx} = 4x
    x = 2 \implies \frac{dy}{dx} = 4(2) = 8
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    (Original post by morgan8002)
    \frac{dy}{dx} = 4x
    x = 2 \implies \frac{dy}{dx} = 4(2) = 8
    yeah?
    so you're saying this 8 is our new gradient? or just value of x * gradient?
    I would have thought it is the result of
    y=4x+c
    we know x =2
    and thus y=8 +c

    but you're saying its
    y=8x+c
    so infact
    17=16+c ?
    c=1
    y=8x +1
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    (Original post by Danny.L)
    yeah?
    so you're saying this 8 is our new gradient? or just value of x * gradient?
    I would have thought it is the result of
    y=4x+c
    we know x =2
    and thus y=8 +c

    but you're saying its
    y=8x+c
    so infact
    17=16+c ?
    c=1
    y=8x +1
    The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

    Yes, y = 8x + 1.
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    (Original post by morgan8002)
    The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

    Yes, y = 8x + 1.
    Yeah, I've been doing this for 5 hours straight now so am getting groggy. It's just clicked.
    The general dx/dy gives you the equation of any goven point, however the tangent isn't any given point, it is a singular value, in this case x=2 y=17
    so we have to then specialise (if that's correct), the dx/dy differentiation to that specific point. I completely get it now!
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    (Original post by morgan8002)
    The differential gives you the general equation for the gradient at any point on the line. Substituting a value of x gives the gradient at that point.

    Yes, y = 8x + 1.
    Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
    y=1/x^3
    x= -1
    I understand that we need to do
    y = 1/-1^3
    and thus y = -1
    so our coordinates are (-1,-1)
    then we differentiate to get
    dx/dy
    so:
    1 * x^-3
    -3 * x-4
    then we have to times put -1 (x) into the -3/ x^4
    where we get -3/1 so -3
    and thus
    (y) -1 = -3 +c
    so c= 2?
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    (Original post by Danny.L)
    Yeah, I've been doing this for 5 hours straight now so am getting groggy. It's just clicked.
    The general dx/dy gives you the equation of any goven point, however the tangent isn't any given point, it is a singular value, in this case x=2 y=17
    so we have to then specialise (if that's correct), the dx/dy differentiation to that specific point. I completely get it now!
    Yes I'm glad you got it. The differential is general to the line. The gradient of the tangent is specific to that point on the line.
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    (Original post by Danny.L)
    Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
    y=1/x^3
    x= -1
    Rearrange to get y =x^{-3}, then follow roughly the same method. Tell me if you get stuck.
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    (Original post by morgan8002)
    Rearrange to get y =x^{-3}, then follow roughly the same method. Tell me if you get stuck.
    Just updated my last reply to you, take a look and see if that looks right to you
    I'm glad I got it too! Thanks for that btw
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    (Original post by Danny.L)
    Just updated my last reply to you, take a look and see if that looks right to you
    I'm glad I got it too! Thanks for that btw
    Yeah it looks right.
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    (Original post by Danny.L)
    Btw, last question, I have to find the equation of a tangent to 1/(x^3), was wondering if you could help?
    y=1/x^3
    x= -1
    I understand that we need to do
    y = 1/-1^3
    and thus y = -1
    so our coordinates are (-1,-1)
    then we differentiate to get
    dx/dy
    so:
    1 * x^-3
    -3 * x-4
    then we have to times put -1 (x) into the -3/ x^4
    where we get -3/1 so -3
    and thus
    (y) -1 = -3 +c
    so c= 2?
    You are almost there
    Your point is (-1,-1)
    Your m is -3

    So y=-3x+c

    Put your point into that and you will see that c=-4
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    (Original post by morgan8002)
    Yeah it looks right.
    Not quite
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    (Original post by TenOfThem)
    You are almost there
    Your point is (-1,-1)
    Your m is -3

    So y=-3x+c

    Put your point into that and you will see that c=-4
    but now I am confused again.
    to get m = -3 I had to put the x=-1 value into it
    1 * x^-3 is the raw form
    differentiated is:
    -3 * x^-4
    so differentiated is
    -3/x^4
    then we make it specific by replacing x with -1
    this gives us
    -3/-1^4
    and thus -3/1 = -3
    so surely our sepcific gradient for the tangent is -3, we dont need to times it by -1 again.
    -3 / x^4 is the gradient of any given point on the curve.
    Replace x with -1 and surely the gradient isn't
    -3x it is just, -3 ? Do you see what I mean? :/
 
 
 
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