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question: find the points at which the graph of x^2 -4x + y62 -6y =12 has a vertical or horizontal tangent line.

I found the deriviative to be (x-2)/(3-y)

Then, the x value for a horizontal would be x=2 and the y value for a vertical would be y=3.

But how do i find the exact points. My teacher gave me these answers

Horizontal tangent at (2,8) and (2,-2)
Vertical tangent at (7,3) and (-3,3)

But i don't understand how she got those points. can someone show me?

Thanks

  • math -

    You've done the hard part, so you're probably just over-thinking it! :-)

    When x=2, plug 2 into that equation of the circle, and what do you get?

    x^2 -4x + y^2 -6y =12

    4 - 8 + y^2 -6y = 12

    y^2 -6y -16 = 0

    and solve for y.

    Same method works for getting the points where y=3.

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