Curve Sketching!! (1 Viewer)

[cutemouse]

Hello,

I'm having trouble sketching y=1/(x²-16)

I find that there are points of inflexion with y''=0... But they seem irrelevant (they're not shown in answers).

I'd appreciate it if someone could help me.

Thanks

 

[tommykins]

vertical ass. at x = +-4
horizontal ass. at y = 0
all positive for x < -4 and x > 4
all negative for -4 < x < 4
y intercept at -1/16

 

[Timothy.Siu]

if there are, its not important

 

[Trebla]

Hello,

I'm having trouble sketching y=1/(x²-16)

I find that there are points of inflexion with y''=0... But they seem irrelevant (they're not shown in answers).

I'd appreciate it if someone could help me.

Thanks

There are NO points of inflexion. Observe:
y = 1 / (x² - 16)
dy/dx = - 2x / (x² - 16)²
u = x, u' = 1
v = (x² - 16)², v' = 4x(x² - 16)
d²y/dx² = - 2 [(x² - 16)² - 4x²(x² - 16)] / (x² - 16)⁴
Any points of inflexion occur when d²y/dx² = 0
- 2 [(x² - 16)² - 4x²(x² - 16)] / (x² - 16)⁴ = 0
(x² - 16)² - 4x²(x² - 16) = 0
(x² - 16)[(x² - 16) - 4x²] = 0
(x² - 16)(3x² + 16) = 0
3x² + 16 = 0 gives no real solution
x² - 16 = 0, gives x = ±4
But the function does not exist at x = ±4
Therefore there are no points of inflexion...

 

[cutemouse]

horizontal ass. at y = 0

Thanks, but how do you get the horizontal ass. ?

all positive for x < -4 and x > 4
all negative for -4 < x < 4

How do you get this bit?

Thanks again, much appreciated!

 

[tommykins]

y=1/(x^2-16)

y =/= 0 as there is a 1 there.

also using limits.

 

[cutemouse]

Thanks mate, how about my 2nd question?

 

[tommykins]

Thanks, but how do you get the horizontal ass. ?



How do you get this bit?

Thanks again, much appreciated!

Sub in extremities, ie. -4.000001 and 4.000001 and you'll see that it starts at a very high number.

since the assymtote is at y = 0, it'll be all positive.

 

[cutemouse]

Hello again,

One last question... for another one.

For the curve y=x²/(x²-9), are there any horizontal asymptotes?

Because as the lim x-->+/- infinity y = 1 ... Did I do something wrong? Answers appears to show that there's a horizontal asym. at y=0, but y can equal zero, when x=0, so...?

BTW: thanks for answering that question!

Thank again!

 

[Trebla]

Horizontal asymptote exists at y = 1.

 

[cutemouse]

Okay thanks...

Then answers are wrong, not suprisingly for Maths In Focus.