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The line y=2x+b is tangent to the graph of y= sqrt(x) at the point P = (a, sqrt(a) ). Find P and determine b.
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|dw:1360027646314:dw| \[2x+b=\sqrt{x}\] \[(2x+b)^2=x\] \[4x^2+4xb+b^2=x\] \[4x^2+x(4b-1)+b^2=0\] Discriminant. \[b^2-4ac=0\] \[(4b-1)^2-4(4)(b^2)=0\] \[16b^2-8b+1-16b^2=0\] \[8b=1\] \[b=\frac{1}{8}\]
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