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3.3 · Q3

Q.Find the equations of the tangents to the curve at points where the tangents to the curve y=2x3−15x2+36x−21y = 2x^3 - 15x^2 + 36x - 21 are parallel to x-axis.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

A tangent is parallel to the x-axis where dydx=0\dfrac{dy}{dx}=0; solve for xx, get the yy-values, and the tangents are horizontal lines y=consty=\text{const}.

Tangent parallel to x-axis   ⟺  dydx=0\iff \dfrac{dy}{dx}=0. Such a tangent has equation y=y0y=y_0 (a horizontal line through the point).

  1. Given y=2x3−15x2+36x−21y=2x^3-15x^2+36x-21, differentiate: dydx=6x2−30x+36\dfrac{dy}{dx}=6x^2-30x+36.
  2. Set the slope to zero: 6x2−30x+36=0⇒x2−5x+6=06x^2-30x+36=0\Rightarrow x^2-5x+6=0.
  3. Factor: (x−2)(x−3)=0⇒x=2(x-2)(x-3)=0\Rightarrow x=2 or x=3x=3.
  4. At x=2x=2: y=2(8)−15(4)+36(2)−21=16−60+72−21=7y=2(8)-15(4)+36(2)-21=16-60+72-21=7, giving point (2,7)(2,7).
  5. At x=3x=3: y=2(27)−15(9)+36(3)−21=54−135+108−21=6y=2(27)-15(9)+36(3)-21=54-135+108-21=6, giving point (3,6)(3,6).
  6. The horizontal tangents are y=7y=7 (at (2,7)(2,7)) and y=6y=6 (at (3,6)(3,6)).
✓Final answer

Points: (2,7)(2,7) and (3,6)(3,6). Tangent equations: y=7y=7 and y=6y=6.

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