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anonymous
 one year ago
Find all points where (4x^2)+(y^2)=8x+4y+4 has a horizontal or vertical tangent line. I graphed it and know that it has horizontal tangent lines at (1,0) and (1,4) and vertical tangent lines at (0,2) and (2,2), I just need to show work.
anonymous
 one year ago
Find all points where (4x^2)+(y^2)=8x+4y+4 has a horizontal or vertical tangent line. I graphed it and know that it has horizontal tangent lines at (1,0) and (1,4) and vertical tangent lines at (0,2) and (2,2), I just need to show work.

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DanJS
 one year ago
Best ResponseYou've already chosen the best response.1have you learned the chain rule , or implicit differentiation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes I have, how would it apply here?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1you want to find the derivative of y with respect to x dy/dx that will tell you the slope of a tangent line to that curve at any point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get that y' is (8x+8)/(2y+4)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1just was looking , i didnt calculate it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no! my original question is a typo, its supposed to be (4x^2)+(y^3)8x+4y+4

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1ha, it's fine, same process

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I get 8x+2y*y'8+4y'=0 and then i isolate y'

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1yeah you just messed up a sign in the result, move all terms with y ' to one side, the rest the other, factor out y', divide by the rest y ' = (4x + 4)/ (y  2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when you take the 2 out does it disappear?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1yeah, just put it in simpler form, factor out 4 in the top and a 2 in the bottom, then simplify that fraction

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1that is the expression for the slope of a tangent to the relationship

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you cant factor out a 2 in the bottom, unless you want to get y/2. so now i have 4(x+1)/y2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1ha, not again, i did once when it was 2y4  2(y2), the 2 divides into the 8 factored from the top

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha i realized that after i posted that. so once i have that how do i find the horizontal and vertical tangent lines?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1do you know the slope of horizontal or vertical lines?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0horizontal is 0 and vertical is undefined, right?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1yep, find out when the derivative are those values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused, how would i do that?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1y ' is the derivative you calculated aka the instantaneous slope of the graph at a point. Set the expression y ' = 0 and solve where that occurs

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1y ' = 0 = (4x + 4) / (y2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0am I solving for x or y

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1just notice you have a fraction, a fraction is zero when the numerater is zero, and the fraction is undefined if the denominator is zero

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1solving, the derivative y ' = 0, when x=1 and undefined at y = 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so how does this get me the 4 points I need?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1That tells you horizontal tangent(s) happen when x=1 vertical tangents(s) happen when y = 2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1use x=1 in the original problem and calculate the y values. 2 possibly since it is a quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and shouldnt it be y=2?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the derivative y ' has y  2 in the bottom, that was what you missed at first, just a sign

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when I solved with y=2 I get the correct points (the same points as when I graphed it with desmos)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1THe tangent is horizontal when x = 1, solving for the y values , you get 2 of them from the quadratic.. \[(1 , ~[2\sqrt{3}+2])~~~~~and~~~~~(1,~ [2\sqrt{3}+2])\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1The undefined value of y = 2 for the vertical tangent points, results in 2 x values from the problem because x is also a quadratic...
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