Find the point at which the tangent to the curve y = 4-x²
Find the point at which the tangent to the curve $y = 4-x^{2}$ is parallel to the chord $AB$, where $A(-2,0)$ and $B(1,3)$.
Explanation of Method
The mean value point is where the tangent matches the chord.
Step 1: Slope of the chord
Both points lie on the curve: $4-4 = 0$ and $4-1 = 3$. Their chord has slope
\[ k = \frac{3-0}{1-\bigl(-2\bigr)} = \frac{3}{3} = 1. \]
Step 2: Match the derivative to it
\[ y' = -2x = 1 \;\Longrightarrow\; \xi = -\frac12. \]
Step 3: The point of tangency
\[ y\left(-\frac12\right) = 4-\frac14 = \frac{15}{4}, \qquad T = \left(-\frac12,\ \frac{15}{4}\right). \]
Step 4: The tangent line
\[ y-\frac{15}{4} = 1\cdot\left(x+\frac12\right) \;\Longrightarrow\; y = x+\frac{17}{4}. \]
Step 5: Verification
\[y'\left(-\frac12\right) = 1 = k \;\;\checkmark; \qquad T \text{ lies on the curve: } 4-\tfrac14 = \tfrac{15}{4} \;\;\checkmark\]
\[\text{The chord is } y = x+2; \text{ the tangent } y = x+\tfrac{17}{4} \text{ is parallel to it, } \tfrac94 \text{ higher} \;\;\checkmark\]
\[\xi = -\tfrac12 \in \bigl(-2,1\bigr), \text{ exactly as the mean value theorem predicts} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,T = \left(-\dfrac12,\ \dfrac{15}{4}\right), \qquad \text{tangent: } y = x+\dfrac{17}{4}\,}\]
Remark
Remark. For a parabola the mean value point is always the midpoint of the interval: here $\tfrac{-2+1}{2} = -\tfrac12$. That is a special feature of quadratics, whose derivative is linear, and it is the reason the tangent at the midpoint of any chord of a parabola is parallel to that chord — a fact known to Archimedes.