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The curve C has the equation y=(x-a)^2, where a is constant. Given that dy/dx = 2x-8 find the value of a
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HI!! make \[x^2-8x\] a perfect square
So, you got that from integrating dy/dx? and hello :)
So \[(x-4)^2 + 8\]?
i would say \[(x-4)^2=x^2-8x+16\] so yeah
actually not what you said, they are only asking you to make it \((x-a)^2\) so the answer is \((x-4)^2\)
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So a would = 4 or -4?
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