Any tangent line to \(y=x^2-x\) at a point \((a,b)=(a,a^2-a)\) will have slope \(y'(a)=2x-1\bigg_{x=a}=2a-1\). Such a line has equation
\(y-(a^2-a)=(2a-1)(x-a)\implies y=(2a-1)x-a^2\)
For this (these) line(s) to pass through (2, -3), you must have
\(-3=(2a-1)\times2-a^2\implies a^2-4a-1=0\implies a=2\pm\sqrt5\)
So the two lines in questions are \(y=(2a-1)x-a^2\) for these values of \(a\).
To show there is no tangent line that passes through (2, 7), you have to show there is no solution for \(a\) in
\(7=(2a-1)\times2-a^2\)
You have
\(a^2-4a+9=0\)
but this has no real solutions, so no such tangent line exists.
