Mathematics
OpenStudy (anonymous):

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)

OpenStudy (anonymous):

sorry what it has to do with the question?

OpenStudy (amistre64):

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

OpenStudy (amistre64):

[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)

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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?

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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)

OpenStudy (amistre64):

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