Question

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Find an equation for the tangent line at the point
(5, 1).
x^2+5y^2-4x+2y=0

Answer 1

To find the tangent line, you have to find the derivative of the equation.

Can you find the derivative of x^2 + 5y^2 - 4x + 2y ?

Answer 2

yes one moment

Answer 3

well with implicit differentiation i get y'=-(x-2)/5y+1 is this correct

Answer 4

or its 2x-4

Answer 5

I didn't do Implict differentiation, and I got:

I got 2x + 10y - 4 + 2

= 2x + 10y - 2

You may be more familier with this than me. Why would you use implicit differentiation.

Answer 6

nono you are differentiating with respect to one variable

Answer 7

you must use implicit

Answer 8

Thats what i did

Answer 9

did you see my answer? is it correct?

Answer 10

Lol listen to zz

Answer 11

x^2+5y^2-4x+2y=0

2x+10yy'-4+2y'=0
10yy'+2y'=-2x+4
y'(10y+2)=-2x+4
y'=(-2x+4)/(10y+2)
y'=2/2(2-x)/(5y+1) = (2-x)/(5y+1)

Answer 12

so yes you are right

Answer 13

okay so would my answer be my final answer?

Answer 14

now plug in the point to get slope, then use y=mx+b

Answer 15

it would be -3/6 ?

Answer 16

if you plug in (5,1) you get (2-5)/(5+1) -3/6=-1/2
so you have
y=mx+b
y = (-1/2)x+b
now use the point again to find b
1=(-1/2)5+b
1=-5/2+b
b = 1+5/2 = 7/2
so
y = (-1/2)x+(7/2)

Answer 17

oh, okay, I'm so into this extreme calculus anymore i forget the basics. Okay, thank you so much! totally understand this a lot more now!

Answer 18

np, as long as you get it now:)