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have you learned the chain rule , or implicit differentiation

yes I have, how would it apply here?

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'

when you take the 2 out does it disappear?

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

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 ' = 0 = (-4x + 4) / (y-2)

am I solving for x or y

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)