Question

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.

Answer 1

have you learned the chain rule , or implicit differentiation

Answer 2

yes I have, how would it apply here?

Answer 3

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

Answer 4

okay one second.

Answer 5

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

Answer 6

umm

Answer 7

is that wrong?

Answer 8

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

Answer 9

just was looking , i didnt calculate it

Answer 10

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

Answer 11

there is no = in it

Answer 12

ha, it's fine, same process

Answer 13

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

Answer 14

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)

Answer 15

when you take the 2 out does it disappear?

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

do you know the slope of horizontal or vertical lines?

Answer 22

horizontal is 0 and vertical is undefined, right?

Answer 23

yep, find out when the derivative are those values

Answer 24

I'm confused, how would i do that?

Answer 25

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

Answer 26

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

Answer 27

am I solving for x or y

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

and shouldnt it be y=-2?

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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...