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    Question:
    Find the possible values of a for which y = ax + 1 is a tangent to  y = 3x^{2} -4x + 4 .

    Please look at my working
    I still cant find a find to get to the answers of -10, 2

    please help
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    Attachment 556475556477

    thank you in advance
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    (Original post by bigmansouf)
    Question:
    Find the possible values of a for which y = ax + 1 is a tangent to  y = 3x^{2} -4x + 4 .

    Please look at my working
    I still cant find a find to get to the answers of -10, 2

    please help
    Name:  Scan0030.jpg
Views: 47
Size:  195.0 KB
    Attachment 556475556477

    thank you in advance
    If b=-(4+a)

    then b^2 = \left[-(4+a)\right]^2 = (4+a)^2

    Just like e.g. (-3)^2 = 3^2
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    (Original post by bigmansouf)
    Question:
    Find the possible values of a for which y = ax + 1 is a tangent to  y = 3x^{2} -4x + 4 .

    Please look at my working
    I still cant find a find to get to the answers of -10, 2
    When you were finding b^2, you made a mistake.
    I think you should expand the brackets first, - (4+a) first = -4 + 4a and then square them.
    Or, do exactly what notnek said - think of it like that, if you find it easier - rearrange the 4 + a to a + 4 so it matches (x+a)^2 and solve it like that.
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    (Original post by Chittesh14)
    When you were finding b^2, you made a mistake.
    I think you should expand the brackets first, - (4+a) first = -4 + 4a and then square them.
    Or, do exactly what notnek said - think of it like that, if you find it easier - rearrange the 4 + a to a + 4 so it matches (x+a)^2 and solve it like that.
    (Original post by notnek)
    If b=-(4+a)

    then b^2 = \left[-(4+a)\right]^2 = (4+a)^2

    Just like e.g. (-3)^2 = 3^2
    thank you i looked through my work and realised i did a mistake
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    (Original post by bigmansouf)
    thank you i looked through my work and realised i did a mistake
    No problem .
 
 
 
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