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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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.

Mathematics
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have you learned the chain rule , or implicit differentiation
yes I have, how would it apply here?
you 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

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okay one second.
I get that y' is (-8x+8)/(2y+4)
umm
is that wrong?
8x + 2y * y' = 8 + 4y '
just was looking , i didnt calculate it
oh no! my original question is a typo, its supposed to be (4x^2)+(y^3)-8x+4y+4
there is no = in it
ha, it's fine, same process
so I get 8x+2y*y'-8+4y'=0 and then i isolate y'
yeah 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)
when you take the 2 out does it disappear?
yeah, just put it in simpler form, factor out 4 in the top and a 2 in the bottom, then simplify that fraction
that is the expression for the slope of a tangent to the relationship
you cant factor out a 2 in the bottom, unless you want to get y/2. so now i have 4(-x+1)/y-2
ha, not again, i did once when it was 2y-4 --- 2(y-2), the 2 divides into the 8 factored from the top
haha i realized that after i posted that. so once i have that how do i find the horizontal and vertical tangent lines?
do you know the slope of horizontal or vertical lines?
horizontal is 0 and vertical is undefined, right?
yep, find out when the derivative are those values
I'm confused, how would i do that?
y ' 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
y ' = 0 = (-4x + 4) / (y-2)
am I solving for x or y
just notice you have a fraction, a fraction is zero when the numerater is zero, and the fraction is undefined if the denominator is zero
solving, the derivative y ' = 0, when x=1 and undefined at y = 2
so how does this get me the 4 points I need?
That tells you horizontal tangent(s) happen when x=1 vertical tangents(s) happen when y = 2
use x=1 in the original problem and calculate the y values. 2 possibly since it is a quadratic
and shouldnt it be y=-2?
the derivative y ' has y - 2 in the bottom, that was what you missed at first, just a sign
when I solved with y=-2 I get the correct points (the same points as when I graphed it with desmos)
THe tangent is horizontal when x = 1, solving for the y values , you get 2 of them from the quadratic.. \[(1 , ~[2\sqrt{3}+2])~~~~~and~~~~~(1,~ [-2\sqrt{3}+2])\]
The 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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