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Since the curves are tangent, they have the same slope at that point and the same x and y value at that point.

Set equations equal and set slopes equal and solve

Togetslope, you need to know calculus first derivative

slope of one equation is dy/dx = 8x +k

and the othereqaution slope is dy/dx = 1

so youhave

4x^2 + kx + 1 = x -8

8x + k = 1

4x^2 +kx -x + 9 = 0

8x + k -1 = 0

solve for k

k ==-11 or +13

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Q: What are the possible values of k for which the line of y equals x -8 is tangent to the curve of y equals 4x squared plus kx plus 1 showing work?
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