## 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.

1. DanJS

have you learned the chain rule , or implicit differentiation

2. anonymous

yes I have, how would it apply here?

3. DanJS

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

4. anonymous

okay one second.

5. anonymous

I get that y' is (-8x+8)/(2y+4)

6. DanJS

umm

7. anonymous

is that wrong?

8. DanJS

8x + 2y * y' = 8 + 4y '

9. DanJS

just was looking , i didnt calculate it

10. anonymous

oh no! my original question is a typo, its supposed to be (4x^2)+(y^3)-8x+4y+4

11. anonymous

there is no = in it

12. DanJS

ha, it's fine, same process

13. anonymous

so I get 8x+2y*y'-8+4y'=0 and then i isolate y'

14. DanJS

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)

15. anonymous

when you take the 2 out does it disappear?

16. DanJS

yeah, just put it in simpler form, factor out 4 in the top and a 2 in the bottom, then simplify that fraction

17. DanJS

that is the expression for the slope of a tangent to the relationship

18. anonymous

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

19. DanJS

ha, not again, i did once when it was 2y-4 --- 2(y-2), the 2 divides into the 8 factored from the top

20. anonymous

haha i realized that after i posted that. so once i have that how do i find the horizontal and vertical tangent lines?

21. DanJS

do you know the slope of horizontal or vertical lines?

22. anonymous

horizontal is 0 and vertical is undefined, right?

23. DanJS

yep, find out when the derivative are those values

24. anonymous

I'm confused, how would i do that?

25. DanJS

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

26. DanJS

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

27. anonymous

am I solving for x or y

28. DanJS

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

29. DanJS

solving, the derivative y ' = 0, when x=1 and undefined at y = 2

30. anonymous

so how does this get me the 4 points I need?

31. DanJS

That tells you horizontal tangent(s) happen when x=1 vertical tangents(s) happen when y = 2

32. DanJS

use x=1 in the original problem and calculate the y values. 2 possibly since it is a quadratic

33. anonymous

and shouldnt it be y=-2?

34. DanJS

the derivative y ' has y - 2 in the bottom, that was what you missed at first, just a sign

35. anonymous

when I solved with y=-2 I get the correct points (the same points as when I graphed it with desmos)

36. DanJS

37. DanJS

THe tangent is horizontal when x = 1, solving for the y values , you get 2 of them from the quadratic.. $(1 , ~[2\sqrt{3}+2])~~~~~and~~~~~(1,~ [-2\sqrt{3}+2])$

38. DanJS

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...