Anonymous
Anonymous asked in Education & ReferenceHomework Help · 4 weeks ago

Find the slope int. equation of the normal line to the graph f(x)=x^3-x^2 at the point where x=1?

1 Answer

Relevance
  • 4 weeks ago
    Favourite answer

    first find the slope of the tangent line by taking the derivative:

    f(x) = x^3 - x^2

    f'(x) = 3x^2 - 2x

    f'(1) = 3 - 2 = 1 <<<====slope of tangent line where x=1

    the normal line is perpendicular to the tangent, so the slope of the normal line is -1.

    find the y-coordinate:

    f(1) = 1 - 1 = 0

    Now you have the slope, m, and point (1,0) and can use that to find the equation:

    y = mx + b

    0 = (-1)(1) + b

    b = 1

    equation in slope-intercept form:

    y = -1x + 1

Still have questions? Get answers by asking now.