A curve with the equation $x^2-2x+3$ has point b, what is the absis of point b if the tangent line on b passes point (1,1)
I solved it by first finding the gradient of the curve which is 2x-2, then I continued making a line equation which passes trough point (1,1)
I got $(2x-2)(x-1)-1$ and because it is the tangent line, and i have to find the absis of point b, my logic is that it must be the intersection. so I did this
$$(2x-2)(x-1)-1=x^2-2x+3$$
which gave me an $x$ of $2$ and $0$.
Am I doing it right just plugging in the unknown gradient like that? if I'm wrong please correct me
I guess you're asking for the two tangents to $y=x^2-2x+3$ passing through $(1,1)$.
You're approach can't be right because you get a polynomial of degree $2$ as equation of a tangent. A expression in the form $y=x^2+\dots$ cannot be a straight line...
I think the standard approach is the following.
A generic (non vertical) straight line passing through $(1,1)$ has the expression
$$y = m(x-1) + 1$$ for some real $m$. You are looking for those values of $m$ which make your line a tangent to $y=x^2-2x+3$.
A tangent of a parabola has the peculiarity that cross the parabola in a single point. The points $x_0$ in which your line cross the parabola must satisfy the two conditions $$y_0 = m(x_0-1) + 1\,;\quad y_0 = x_0^2-2x_0+3\,.$$ This subtracting the two expressions you get that $x$ must solve $$x_0^2-(2+m)x_0+2+m=0\,.$$ You want that the parabola and the line cross in a single point to have a tangent. So you want that the above equation has a single solution for $x_0$.
This means that the discriminant $\Delta$ of the second order equation must be $0$. You have
$$\Delta = (m-2)^2-4(m+2)=(m-2)(m+2),.$$
So your solution are $m=2$ and $m=-2$.
Indeed the two tangents to your parabola passing through $(1,1)$ are $$y = 2(x-1) + 1 = 2x-1$$ and $$y = -2(x-1)+1 = -2x+3\,.$$
Now the coordinate of $b$ can be find easily. You get that the two possibilities for $b$ are $(0,2)$ and $(2,2)$.