Question

sketch graph of function f(x)=x^2/(x^2-4) using any extrema, intercepts, symmetry, and asymptotes

Answer 1

ok, well you are going to need the first and second derivatives , so why not start out by finding those

Answer 2

yeah i know, currently finding both the derivatives

Answer 3

\[\frac{ d }{ dx }\frac{ x^2 }{ (x^2-4) }\]

Answer 4

\[\frac{ -8 }{ (x ^{2}-4 )^{2}}\]

Answer 5

*-8x

Answer 6

From the denominator of f(x), it factors to
(x+2)(x-2), so if the denominator is zero, the function does not exist,
for X=2 and X=-2 you have
vertical asymptotes

Answer 7

\[\lim_{x \rightarrow +- \infty} f(x) = \]

Answer 8

As x goes off to infinity or negative infinity, the function f(x) goes to 1
Horizontal asmytope at Y=1

Answer 9

\[\lim_{x \rightarrow \infty} \frac{ 1 }{ 1 - \frac{ 4 }{ x^2 } }\]

Answer 10

divided everything by x^2 , then let x go to infinity,

Answer 11

oh k, i already know of the asmytopes, jut having trouble on finding the derivatives, and the intercepts

Answer 12

ok...

Answer 13

mostly the second deriv

Answer 14

The first derivative is missing a x term in the numerator,

\[y ' = \frac{ -8x }{ (x^2-4)^2 }\]

Answer 15

using the quotient rule

Answer 16

oh i see you corrected it

Answer 17

so

\[y '' = \frac{ d }{ dx }\frac{ -8x }{ (x^2-4)^2 }\]

Answer 18

same quotient rule as the first derivative gives...

\[y '' = \frac{ 8(3x^2 +4) }{ (x^2 - 4)^2 }\]

Answer 19

yeah i keep ending up a ridiculous number as an answer for the second derivative, such that i'm struggling on it

Answer 20

oh k thank you

Answer 21

you could let
u= x^2 + 4
du = 2x dx

Answer 22

oh yeah, i forgot that i could do that huh

Answer 23

ok so now you need to test for
Increasing/Decreasing
Concave up/Concave Down

Answer 24

i know that the first one is the critical of 0, 2

Answer 25

well 2 is a vertical asmtope

Answer 26

That is what we have so far, with x=0 a critical point

Answer 27

oh k, then wouldn't there be zero inflection?