# Find the y -intercept of the curve that passes through the point ( 2 , 1 ) wi

Find the $y$-intercept of the curve that passes through the point $\left(2,1\right)$ with the slope at $\left(x,y\right)$ of $\frac{-9}{{y}^{2}}$
$\frac{dy}{dx}=\frac{-9}{{y}^{2}}$
$\int {y}^{2}dy=\int -9dx$
$\frac{{y}^{3}}{3}=-9x+{C}_{1}$
${y}^{3}=-27x+C$ $\left(C=3{C}_{1}\right)$
$y=\left(-27x+C{\right)}^{1/3}$
$1=\left(-27\left(2\right)+C{\right)}^{1/3}$
$C=54$
$0=\left(-27x+54{\right)}^{1/3}$
$y-intercept=\left(-2,0\right)$
Where is mistake?
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Aibling6n2re
You've miscalculated when finding your $C.$ When plugging $\left(x,y\right)=\left(2,1\right)$ into the equation ${y}^{3}=-27x+C$ (there's no need to solve for $y$ first, since the basic cubic function is invertible), we get
$1=-54+C,$
Furthermore, you went on to use $54$ as your $C$ value, though you allegedly found $C=52.$ I suspect that at least one typo has been made.
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