Show that the tangents to the curve y = 7x^3 + 11 at the points where x = 2 and x = −2 are parallel.
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we have to prove : the tangents to the curve y = 7x³ + 11 at the point where x = 2 and x = -2 are parallel.
y = 7x³ + 11 , differentiate with respect to x,
dy/dx = 21x²
now, slope of tangent of the curve at (x = 2) =
dy/dx at (x = 2) = 21(2)² = 84
again, slope of tangent of the curve at (x= -2) = dy/dx at (x = -2) = 21(-2)² = 84
e.g., slope of tangent at (x = 2) =slope of tangent at (x = -2)
we know, two lines will be parallel only when slope of the given lines must be equal.
hence, it is proved that tangent to the curve y = 7x³ + 1 at the points where x = 2 and x= -2 are parallel.
y = 7x³ + 11 , differentiate with respect to x,
dy/dx = 21x²
now, slope of tangent of the curve at (x = 2) =
dy/dx at (x = 2) = 21(2)² = 84
again, slope of tangent of the curve at (x= -2) = dy/dx at (x = -2) = 21(-2)² = 84
e.g., slope of tangent at (x = 2) =slope of tangent at (x = -2)
we know, two lines will be parallel only when slope of the given lines must be equal.
hence, it is proved that tangent to the curve y = 7x³ + 1 at the points where x = 2 and x= -2 are parallel.
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