Q.9 Find the points on the curve given by y=x³-6x²+x+3 where the tangents are parallel to the line y=x+5 .

Answer: (0,3) and (4,-25) Slope of the tangent: dy/dx=3x²-12x+1 . The slope of y=x+5 is 1. For parallel tangents: 3x²-12x+1=1⇒3x(x-4)=0⇒ x=0 or x=4 . When x=0 , y=3 ; when x=4 , y=64-96+4+3=-25 . The required points are (0,3) and (4,-25) .

Chapter: Applications of Derivatives· Mathematics & Statistics Part 2

Find the points on the curve given by \(y=x^3-6x^2+x+3\) where the tangents are parallel to the line \(y=x+5\).

Q.9 Find the points on the curve given by \(y=x^3-6x^2+x+3\) where the tangents are parallel to the line \(y=x+5\).

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \((0,3)\) and \((4,-25)\)

Slope of the tangent: \(\dfrac{dy}{dx}=3x^2-12x+1\). The slope of \(y=x+5\) is 1.

For parallel tangents: \(3x^2-12x+1=1\Rightarrow3x(x-4)=0\Rightarrow x=0\) or \(x=4\).

When \(x=0\), \(y=3\); when \(x=4\), \(y=64-96+4+3=-25\).

The required points are \((0,3)\) and \((4,-25)\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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