Question

Find the equations of the lines tangent to y=x^2 and pass through the point (1,-3).

Answer 1

What I've already gotten it down to is [(x^2-3)/(x-1)]=2x

Answer 2

Sorry, x^2+3

Answer 3

I figured it out.
\[(x^2+3)/(x-1)=2x\]
\[2x^2-2x-x^2-3=0\]
\[(x-3)(x+1)=0\]
\[x=3, -1\]

Answer 4

\[y+3=-2(x-1)\]
\[y=-2x-1\]

Answer 5

\[y+3=6(x-1)\]

\[y=6x-9\]