Find the equation of the tangent to the curve with equation
y=16x3−9x12 y = \frac{1}{6}x^3 - 9x^\frac{1}{2} y=61x3−9x21at the point P(9,94.5)P(9, 94.5)P(9,94.5).
Give your answer in the form ax+by+c=0ax + by + c = 0ax+by+c=0, where aaa, bbb and ccc are integers.
The curve with equation y=f(x)y = f(x)y=f(x) also passes through the point P(9,94.5)P(9, 94.5)P(9,94.5). Given that
f′(x)=16x3−9x12 f'(x) = \frac{1}{6}x^3 - 9x^\frac{1}{2} f′(x)=61x3−9x21find f(x)f(x)f(x), giving the coefficients in simplest form.