Q.10 Show that the tangent to the curve y = x³ - 6x² + x + 3 at the point (0, 3) is parallel to the line y = x + 5 .

Show that the tangent to the curve \(y = x^3 - 6x^2 + x + 3\) at the point \((0, 3)\) is parallel to the line \(y = x + 5\).

Q.10 Show that the tangent to the curve \(y = x^3 - 6x^2 + x + 3\) at the point \((0, 3)\) is parallel to the line \(y = x + 5\).

Doubtly Asked 28 Oct 2026 Board exam · July 2025 hsc 12th Save WhatsApp Report

Best answer

Answer: Both have slope 1, so the tangent is parallel to the line.

\(\frac{dy}{dx} = 3x^2 - 12x + 1\). At \((0, 3)\), slope of tangent \(= 0 - 0 + 1 = 1\).

Slope of the line \(y = x + 5\) is 1. Since the slopes are equal, the tangent at \((0, 3)\) is parallel to the line \(y = x + 5\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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