Calculus derivative

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Find the slope and an equation of the tangent line to the graph of the function f at the specified point.

f(x)=-1/3x^2+5x+5: (-1, -1/3)

Answer: f’(x)=-2/3x + 5

y=-2/3x -1

I re-worked the problem and got f'(x)=5, y=5x+ 14/3

This is from a multiple choice test and this is not an answer.

Can someone check my work?

  • Calculus derivative -

    Your derivative is correct
    Now use the value of x of the given point to find the slope
    slope = (-2/3)(-1) + 5
    = 2/3 + 5 = 17/3

    equation is
    y = (17/3)x + b
    sub in the point
    -1/3 = (17/3)(-1) + b
    16/3 = b

    equation: y = (17/3)x + 16/3

  • Calculus derivative -

    Thanks. I was missing the step where you plug in the -1 to get slope.

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