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    a) Find the equation of the normal to the curve y=e^-3x+2 at the point (0,3), and find the coordinates of the point where this normal crosses the x- axis.

    b)Fine the region bounded by the curve with equation y=e^-3x+2, the coordinate axis and the line x=1.

    Pleaseee help I can't get the answer through my differentiation
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    (Original post by Carolyne2891)
    a) Find the equation of the normal to the curve y=e^-3x+2 at the point (0,3), and find the coordinates of the point where this normal crosses the x- axis.

    b)Fine the region bounded by the curve with equation y=e^-3x+2, the coordinate axis and the line x=1.

    Pleaseee help I can't get the answer through my differentiation
    Then show us your process and we'll show you where you trip up. We're not doing the Q for you.
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    a) Find the derivative and plug in x=0 then find -1/that. Plug that into your point and find the equation. Set this to 0 and solve.
    b) I don't believe integration is C3. However, this may be a thinking problem. If you know dy/dx = y then you know the integral of y must also be y. Use this concept and the definition of an integral (the opposite of a derivative) to solve the equation. Find the definite integral between the given limits.
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    dy/dx= -3e^-3x
    which is over 1/(-3e^3x)
    meaning m of normal is 3e^3x
    y=e^-3x +2
    0=e^-3x +2
    not sure how to solve for x with logs,
    but the answer is (-9,0) and equation 3y=x+9 ???
    Not a clue
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    (Original post by Carolyne2891)
    dy/dx= -3e^-3x
    which is over 1/(-3e^3x)
    meaning m of normal is 3e^3x
    y=e^-3x +2
    0=e^-3x +2
    not sure how to solve for x with logs,
    but the answer is (-9,0) and equation 3y=x+9 ???
    Not a clue
    It asks for the gradient at point (0,3) so you have to take x as 0 to find the gradient. Also, only e has a negative power so if you want to use that form it's -3/e^3x.
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    (Original post by Carolyne2891)
    dy/dx= -3e^-3x
    which is over 1/(-3e^3x)
    meaning m of normal is 3e^3x
    That's not right. \displaystyle -3e^{-3x}=-3\cdot \frac{1}{e^{3x}} \neq \frac{1}{-3e^{3x}}

    So your normal gradient function is \frac{e^{3x}}{3}

    You now want the normal gradient the point where x=0, thus construct the equation of your normal.
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    Thank you everyone!
 
 
 
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