AS LEVEL MATHS EQUATION! Who can solve this? :)

Find an equation of the tangent and the normal at the point where x=2 on the curve with the equation y=8/x-x+3x^2, x>0.

The answer is y=9x-4 and 9y+ x=128
HOWEVER I NEED THE WORKING OUT!! :) thankyou soo much :)

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Just remember:

1) The derivative of the equation with respect to x is the slope of the tangent line.
2) The slope of the normal line is the negative inverse of the slope of the tangent line.

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y=8/x-x+3x^2, x>0

The derivative of 8/x-x+3x^2 is -8/x^2 + 6x - 1.

Evaluate the original equation at x = 2:
y = 8/2 - 2 + 3*4
y = 14

Evaluate the derivative at x = 2:
-8/4 + 12 - 1 = 9

The derivative (9) is the slope of the tangent line. The tangent line goes through the point (2,14).

So, in point slope form the equation of the tangent line is y-14 = 9(x-2), or y = 9x - 4 in slope intercept form.

The slope of the normal line is the negative inverse of the slope of the tangent line. So, the slope of the normal is -1/9. The normal also goes through (2,14), so in point slope form the equation of the normal is
y-14 = (-1/9)(x-2), or y = -x/8 + 128 in slope intercept form.