From the point (2,-2) draw the tangent to the parabola y
From the point $\bigl(2,-2\bigr)$ draw the tangent to the parabola $y = x^{2}-3x+1$ and find the point of tangency.
Explanation of Method
An external point, and therefore two tangents.
Step 1: The tangent at an unknown point
Let $\bigl(x_{0},y_{0}\bigr)$ be the point of contact, so $y_{0} = x_{0}^{2}-3x_{0}+1$ and, since $y' = 2x-3$, the slope is $k = 2x_{0}-3$. The tangent is
\[ y-y_{0} = k\bigl(x-x_{0}\bigr). \tag{1} \]
Step 2: Make it pass through $\bigl(2,-2\bigr)$
\[ -2-\bigl(x_{0}^{2}-3x_{0}+1\bigr) = \bigl(2x_{0}-3\bigr)\bigl(2-x_{0}\bigr), \]
\[ -x_{0}^{2}+3x_{0}-3 = -2x_{0}^{2}+7x_{0}-6, \]
\[ x_{0}^{2}-4x_{0}+3 = 0 \;\Longrightarrow\; \bigl(x_{0}-1\bigr)\bigl(x_{0}-3\bigr) = 0. \tag{2} \]
Step 3: The two tangents
\[ x_{0} = 1:\ y_{0} = -1,\ k = -1 \;\Longrightarrow\; y = -x; \]
\[ x_{0} = 3:\ y_{0} = 1,\ k = 3 \;\Longrightarrow\; y = 3x-8. \]
Step 4: Verification
\[\text{Through } \bigl(2,-2\bigr):\ -2 = -2 \;\;\checkmark; \qquad 3\cdot2-8 = -2 \;\;\checkmark\]
\[\text{On the parabola: } 1-3+1 = -1 \;\;\checkmark; \qquad 9-9+1 = 1 \;\;\checkmark\]
\[\text{Slopes: } y'(1) = -1,\ y'(3) = 3 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y = -x \ \text{ touching at } \bigl(1,-1\bigr); \qquad y = 3x-8 \ \text{ touching at } \bigl(3,1\bigr)\,}\]
Remark
Remark. Two tangents appear because $\bigl(2,-2\bigr)$ lies below the parabola: at $x = 2$ the curve has the value $4-6+1 = -1\gt -2$. A point on the curve would give a repeated root of (2), and a point above the curve none at all — the discriminant of (2), namely $16-12 = 4$, measures exactly how far outside the point lies.