Equation of a tangent line - 2

Geometry Level 1

Find the equation of the line (in slope-intercept form) tangent to the curve y = x 3 − 3 x 2 − 2 y=x^3-3x^2-2 at the point ( 1 , − 4 ) (1,-4) .

y = 3 ( x + 1 ) y=3(x+1) − 1 + 3 x = − y -1+3x=-y 3 x + y = 1 3x+y=1 y = − 3 x − 1 y=-3x-1

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2 solutions

Relevant wiki: Tangent to a Curve

The derivative of a function is identical with the slope of the graph of the function. We have

y = x 3 − 3 x 2 − 2 y=x^3-3x^2-2

y ′ = 3 x 2 − 6 x y'=3x^2-6x

When x = 1 x=1 , we have

y ′ = 3 ( 1 ) 2 − 6 ( 1 ) = − 3 y'=3(1)^2-6(1)=-3

From the point-slope form of an equation of the line, we have

y − y 1 = m ( x − x 1 ) y-y_1=m(x-x_1)

y + 4 = − 3 ( x − 1 ) y+4=-3(x-1)

y + 4 = − 3 x + 3 y+4=-3x+3

y = − 3 x − 1 y=-3x-1

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Edwin Gray
Feb 26, 2019

The line has equation y = mx + b. The slope, m , is dy/dx = 3x^2 - 6x evaluated at x = 1, so m = -3. When x = 1, y = -4, so -4 = (-3)(1) + b, and b = -1. So y = -3x - 1.

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