Find equation of the tangent lines at the points (1, 5) and (2, 3).

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Suresh Chand · Accepted Answer

Solution: Given Equation is: y = 3 + 4x2 - 2x3 First find the derivative of the equation y' = 8x - 6x2 Then find the slope, Slope for (1,5) = m1 = 8 - 6 = 2 Slope for (2,3) = m1 = 16 - 24 = -8 And we will find the equation using y - y1 = m(x - x1) For point (1, 5) y - 5 = 2(x - 1) y - 5 = 2x - 2 y = 2x + 3 For point (2, 3): y - 3 = -8(x - 2) y - 3 = -8x + 16 y = -8x + 19

Find equation of the tangent lines at the points (1, 5) and (2, 3) on the curve y = 3 + 4x² – 2x³

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Find equation of the tangent lines at the points (1, 5) and (2, 3) on the curve y = 3 + 4x2 – 2x3

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Find equation of the tangent lines at the points (1, 5) and (2, 3) on the curve y = 3 + 4x² – 2x³