Determine where, if anywhere, the tangent line to f(x)=x^3−5x^2+x is parallel to the line y=4x+23

  (1/3(5-√34), 1/27(-205+32√34)) and (1/3(5+√34), 1/27(-205-32√34))

Step-by-step explanation:

The slope of the given line is the x-coefficient, 4. Then you're looking for points on the f(x) curve where f'(x) = 4.

  f'(x) = 3x^2 -10x +1 = 4

  3x^2 -10x -3 = 0

  x = (5 ±√34)/3 . . . . . x-coordinates of tangent points

Substituting these values into f(x), we can find the y-coordinates of the tangent points. The desired tangent points are ...

  (1/3(5-√34), 1/27(-205+32√34)) and (1/3(5+√34), 1/27(-205-32√34))

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The graph shows the tangent points and approximate tangent lines.

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