Asked by alison
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
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
Answers
Answered by
jim
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.
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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