y = 3 + 3x

y = 3x + 3, we know that the slope is 3 because the line is in the form y = mx + b, where m is the slope

parallel lines have equal slopes so from the straight line equation the gradient of the tangent must be 3.

y=x sqrt x= x*x^(1/2)=x^(3/2)

dy/dx=3/2x^(1/2)

3=3/2x^(1/2)

2 = sqrt(x) 4 = x

so now we know that the tangent line hits the curve at x = 4. plug that into the original to find the y - value y = 4(sqrt(4)) y = 4(2) = 8

So the point on the curve is (4,8).

Use the linear formula: y-y1=m(x-x1)

y-8=3(x-4)

y-8=3x-12

y=3x-4

answer the tangent line to the curve y = x sqrt x is y=3x-4

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