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

Question

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

2 Answers

Impact of this question

15973 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-24T02:22:19

For the intercepts you set alternately $x = 0$ $x=0$ and $y = 0$ $y=0$ in your function:
enter image source here
and graphically:

Answer 2 · 2015-03-24T02:22:53

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

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

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

Science

Math

Humanities

... and beyond

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

2 Answers

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the intercepts of x^2y-x^2+4y=0x2y−x2+4y=0x^2y-x^2+4y=0?

2 Answers

Related questions

Impact of this question

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