# How do you find the intercepts of x^2y-x^2+4y=0?

Mar 24, 2015

For the intercepts you set alternately $x = 0$ and $y = 0$ in your function:

and graphically:

Mar 24, 2015

On the X-axis $y = 0$
So
${x}^{2} y - {x}^{2} + 4 y = 0$
becomes
${x}^{2} \left(0\right) - {x}^{2} + 4 \left(0\right) = 0$
$\rightarrow - {x}^{2} = 0$
$\rightarrow x = 0$

On the Y-axis $x = 0$
and the original equation
${x}^{2} y - {x}^{2} + 4 y = 0$
becomes
${\left(0\right)}^{2} y - {\left(0\right)}^{2} + 4 y = 0$
$\rightarrow y = 0$

The only intercept for the given equation occurs at $\left(0 , 0\right)$