Question

Prove that the tangents to the curve y = x3 + 6 at the points (-1,5) and (1,7) are parallel.
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Answer 1

SOLUTION

TO PROVE

The tangents to the curve y = x³ + 6 at the points (-1,5) and (1,7) are parallel.

CONCEPT TO BE IMPLEMENTED

For a given curve two tangents at two different point are parallel if they have the same slope

EVALUATION

Here the given equation of the curve is

y = x³ + 6

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = 3 {x}^{2} }$

For the point (-1,5) the slope of the tangent

$\displaystyle \sf{m_1 = \frac{dy}{dx} \bigg|_{( - 1 ,5)} = 3 \times {( - 1)}^{2} = 3}$

For the point (1,7) the slope of the tangent

$\displaystyle \sf{m_2 = \frac{dy}{dx} \bigg|_{( 1 ,7)} = 3 \times {( 1)}^{2} = 3}$

Since

$\displaystyle \sf{m_1 = m_2 }$

Hence the tangents to the curve y = x³ + 6 at the points (-1,5) and (1,7) are parallel.

Hence proved

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