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find k such that the line is tangent to the graph of the function fuction= f(x)=x^2-kx line y=4x-9
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find the 1st derivative..
i did
ok... the gradient/slope of the given line y = 4x - 9 is..?
4
ok what do i do
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so you need to set the 1st derivative equal to 4 and solve for k
i solved for k k=2x-4
ok... so you have \[x^2 - (2x - 4)x\] you need to now find the point of intersection so equate the parabola and line x^2 - (2x - 4)x = 4x - 9 after simplifying -x^2 = -9 x^2 = 9 so x = 3 (3, 3) is the point of intersection.
thanks!
so m = 4 when x = 3 so using the 1st derivative 4 = 2(3) - k k = 2 hope it makes sense... it was a bit more difficult than I thought...
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