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last calc question, i promise!

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given the curve x + xy + 2y^2 = 6...

a. find an expression for the slope of the curve.
i got (-1-y)/(x + 4y) as my answer.
b. write an equation for the line tangent to the curve at the point (2,1).
i got y = (-1/3)x + (5/3).

but i didn't any answer for c!
c. find the coordinates of all other points on this curve with slopes equal to the the slope at (2,1).

do i just set (-1-y)/(x+4y) equal to -1/3?

  • last calc question, i promise! -

    i mean: do i just set (-1-y)/(x+4y) equal to -1/3 AND use the original formula to come up with an answer? like a two-variable system?

  • last calc question, i promise! -

    ALSO ALSO ALSO - on my first calc question (a couple posts down!), i added more to the problem that i forgot, so i need help with that too.

  • last calc question, i promise! -

    a) is correct
    b) correct
    c) yes, set (-1-y)/(x+4y) equal to -1/3
    solving this I got x = 3-y
    sub that back into the original equation to get after simplifying
    y^2 + 2y - 3 = 0
    (y+3)(y-1) = 0
    y = -3 or y = 1
    but those back into x=3-y for
    x =6 or x = 4

    so the other points are (6,-3) and (4,1)

    you better check my math, it is getting late here.

  • last calc question, i promise! -

    the second one would be x = 2, resulting in the original point (2,1), i think. but other than that, thank you sooooo much!

  • last calc question, i promise! -

    No.

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