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Maddie, The derivative gives you the slope of the tangent line at any point (X,Y) on the graph of the function. And then you can use the coordinates of that point to find the equation of the line. You can replace Y by what it equals to get the coordinates in terms of X only. Using the slope of the tangent line at (X,Y) obtained earlier, and remembering how to find the equation of a line given its slope and a point it contains (pointslope form), you can complete the question. Victoria  


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