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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

Best Answer 

 #1
avatar+92673 
+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
avatar+92673 
+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
 #2
avatar+2353 
0

reinout-g  Jun 12, 2014

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