Tangent lines at a point

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  F'(0) = 6*(0)^2 - 8*0 +3 = 3 F'(1) = 6*(1)^2 - 8*1 + 3 = 1 They are the slopes of the tangent lines to these points. E) Find a function for the normal line to the curve at the point (0,-5).

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  According to the point-slope form, the equation of the tangent line is At the two points and , the slope of the tangent line is . So we can say that the tangent lines are not defined at the two points. An astroid is investigated.

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• Mathematics - Calculus - Functions

  Slope of the function at this point is 3(0^2) - 3 = -3. The equation of the tangent line at the point of inflection (0, -1) is y + 1 = -3(x - 0)  y = -3x - 1.

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  The function Has a first derivative: The tangent lines are horizontal when the derivative is zero (the derivative at piont x0 is the slope of the tangent line to the function at x0).

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• Gradients, Derivatives, Tangent Lines, Trajectory and Rates of Change

  115040 Gradients, Derivatives, Tangent Lines, Trajectory and Rates of Change 1) A particle is moving in R^3 so that at time t its position is r(t) = (6t, t^2,t^3). a.