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The tangent to y = x² at the point (1, 1) has slope:
A$1$ (the y-value at x = 1)
B$2$ (since dy/dx = 2x, at x = 1)
C$0.5$ (incorrect inverse calculation)
D$0$ (a horizontal tangent)
Answer & Solution
Correct answer: B. $2$ (since dy/dx = 2x, at x = 1)
dy/dx = 2x. At x = 1: slope = 2(1) = 2. Tangent line: y − 1 = 2(x − 1) → y = 2x − 1. Normal slope = -1/2.
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