How do I find out the x and y intercepts when a graph is shown?

x^4+y^2-xy=16

sally1
Jun 12, 2014

#1**+8 **

This is certainly an unusual one, huh?? To find the x intercept, let's just let y=0

So we have x^4 = 16 → x^4 - 16 = 0 ... which factors as

(x^2 - 4) (x^2 + 4) = 0 .... the second thing has a non-real solution but we can set the frist factor = 0 and find two real solutions

(x^2 - 4) = 0 .. factor as .. (x+2) (x-2) = 0 so the x intercepts occur at (2,0) and (-2,0)

To find the y intercepts, let x = 0....so we have

y^2 = 16 → y^2 - 16 = 0 factor this as .... (y+4) (y-4) =0 and setting each factor to 0 produces y intercepts at (0,4) and (0, -4)

I know that the "xy" term produces the graph of x^4 + y^2 = 16 with a "rotational" element added...heck, I'd even like to look at it myself !!! (If I could add a stem at the top, it might resemble a strange "apple.')

Wow!!!.....that's certainly an odd duck!!...note that the intercepts are right where we said they would be, but as for the rest?? I couldn't begin to guess the shape of this one!!

CPhill
Jun 12, 2014

#1**+8 **

Best Answer

This is certainly an unusual one, huh?? To find the x intercept, let's just let y=0

So we have x^4 = 16 → x^4 - 16 = 0 ... which factors as

(x^2 - 4) (x^2 + 4) = 0 .... the second thing has a non-real solution but we can set the frist factor = 0 and find two real solutions

(x^2 - 4) = 0 .. factor as .. (x+2) (x-2) = 0 so the x intercepts occur at (2,0) and (-2,0)

To find the y intercepts, let x = 0....so we have

y^2 = 16 → y^2 - 16 = 0 factor this as .... (y+4) (y-4) =0 and setting each factor to 0 produces y intercepts at (0,4) and (0, -4)

I know that the "xy" term produces the graph of x^4 + y^2 = 16 with a "rotational" element added...heck, I'd even like to look at it myself !!! (If I could add a stem at the top, it might resemble a strange "apple.')

Wow!!!.....that's certainly an odd duck!!...note that the intercepts are right where we said they would be, but as for the rest?? I couldn't begin to guess the shape of this one!!

CPhill
Jun 12, 2014