Find equations of all lines tangent to the curve at the given value of x

[dubb]

Find equations of all lines tangent to the curve at the given value of x
x + y^2 - y = 1; x = 1

i'm just not sure how to go about doing this, this is my first semester of calculus and i've never taken it before. This problem is in the implicit differentiation section of the book.

so far I think I calculated the derivative:

x + y^2 - y = 1
1 + 2y * y' - y' = 0
y' (2y-1) = -1
y' = -1 / (2y-1)

Thanks for your help.

 

[Ishuda]

What you have done is correct but you could also treat the equation as a quadratic in y and solve for the roots of the equation, i.e.
y² - y + x-1 = 0
or
y = $\displaystyle \frac{1}{2}[ 1 \pm \sqrt{1-4(x-1)} ] = \frac{1}{2}[\space 1 \pm \sqrt{5 - 4 x}\space ]$
So, when x is 1, y is either zero or 1 and y' is either 1 or -1.

Or, maybe easier
y² - y + x - 1 = y (y-1) + x-1 = 0
So at x equals 1
y (y-1) = 0
and y is either zero or 1.

In any case you would have two lines with slopes 1 and -1 going through the points (x,y) = (1, 0) and (1, 1) respectively.

 

[dubb]

Ok, so Im trying to understand how to put it into the form of equations for all tangent lines. Would you be able to help me on this part? Knowing the teacher, I will have to use the derivative because of the lesson that the problem is about.

 

[dubb]

Ok, so Im trying to understand how to put it into the form of equations for all tangent lines. Would you be able to help me on this part? Knowing the teacher, I will have to use the derivative because of the lesson that the problem is about.

Hold, on i think I may have it!

 

[dubb]

OMG, I got it! Thanks, from what you showed me. So this is what I did.
I plugged in x = 1 into the equation x+y^2-y=1, and got y(y-1)=0 which gives me y = 1 and 0
So my points are (1,1) and (1, 0)

to find the slope I use the derivative of the equation:
y' = - 1/(2y-1)

for (1,1) the slope came to - 1
for (1,0) the slope came to 1
so the equations are y = x-1 and y= -x+2 ! that checks out in the book. Thanks again.