What is the equation of the line tangent to f(x) = 5 – x2 at x = –2?

The point on the curve at x = −2 is f(−2) = 5 − (−2)² = 1. The derivative is f′(x) = −2x, so the tangent slope at x = −2 is 4; using point-slope form gives y − 1 = 4(x + 2), choice C.

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