Thread: Finding intercepts?

Can anyone help me find the intercept of this

x^2y-x^2+4y=0

My second question is find the points of intersection of the graphs of the equations
analytically.

x^2+y^2=5

x-y=1

I know you have to set them to y so they can be equal to each other.

So I did y=x-1

But how would I do y^2=-x^2+5 ?

Originally Posted by homeylova223

Can anyone help me find the intercept of this

x^2y-x^2+4y=0 <--- is this written correctly?

My second question is find the points of intersection of the graphs of the equations
analytically.

x^2+y^2=5

x-y=1

I know you have to set them to y so they can be equal to each other.

So I did y=x-1

But how would I do y^2=-x^2+5 ?

In this case I would replace the y of the first equation by x - 1. You'll get a quadratic equation in x:



Solve for x and afterwards plug in the results into the equation y = x - 1 to get the y-values.

Originally Posted by homeylova223

Can anyone help me find the intercept of this

x^2y-x^2+4y=0

For the first one, the graph of this function touches the y-axis when x=0
and touches the x-axis when y=0.

Let one of these equal zero and discover the value of the other.
This is how to find axis crossing-points.

My second question is find the points of intersection of the graphs of the equations
analytically.

x^2+y^2=5

x-y=1

I know you have to set them to y so they can be equal to each other.

So I did y=x-1

But how would I do y^2=-x^2+5 ?

I'd also recommend Earboth's solution.

Hmm I think I may have written it incorrectly

here is what I mean

(x^2)y-(x^2)+4y

Originally Posted by homeylova223

Hmm I think I may have written it incorrectly

here is what I mean

(x^2)y-(x^2)+4y

Same thing if it all equals zero as before.
Just place x=0 and find y
or vice versa.

So If I place y=0 I get x^2(0)-x^2=0

So Would my answer be (0,0)

Yes, that's all there is to it.
The curve touches both axes at the origin.