Question

Find the equations of the tangent lines to the following curves, at the specified
points: ( y = f ( x) )

2

a. y − xy − 6 = 0 at all points satisfying x = 1

3

b. 6 y + 1 − 2 x

32

− 2 = 0 at the point (4, 2)

Answer 1

sorry what it has to do with the question?

Answer 2

the tangent lines; are the slope of the line at any given point on a curve; which can be determined by the derivative of the function

Answer 3

[y − xy − 6 = 0]' = y' -(x'y+xy') = 0

y' -x'y -xy' = 0

y'(1-x) = x'y ; x'=1

y' = y/(1-x)

Answer 4

i cant tell what the question is asking for other than that, the formatting is off

Answer 5

looks right, but i dont understand my steps. what does it mean all points that satisfy x=1?
actually function looks like 2

y^2 − xy − 6 = 0

Answer 6

well, to me that means we set x = 1 and solve for the derivative of the equation:

y^2 -(1)y -6 = 0 is now a normal looking quadratic that can be solved for its zeros

Answer 7

the function is 0 when x = 1 and y = 3,-2 if i see it right

Answer 8

the derivative of the function at these points is:

y' 2y - (x'y + xy') - 0 = 0

y' 2y - x'y - xy' = 0

y' 2y - xy' = x'y

y'(2y - x) = x'y

y' = x'y/(2y - x) ; where x' = 1

y' = y/(2y - x) ; is the slope of the line at (1,3) and (1,-2)

they might be asking you to find the equation of the tangent lines at these points tho?

Answer 9

and they are since I read it again and it says that lol

Answer 10

y = f'(1)x -(f'(1)) + 3 is one of them for (1,3)

y = f'(1)x -(f'(1)) - 2 is the other for (1,-2)

where:

f'(1) = 3/(2(3) - 1) ; at (1,3)

f'(1) = -2/(2(-2) - 1) ; at (1,-2)

Answer 11

hope that makes more sense outside of my head than it did inside of it ;)