anonymous
  • anonymous
Graph 3/x^2-4. Then find the domain/range/VA
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Astrophysics
  • Astrophysics
\[f(x) = \frac{ 3 }{ x^2 }-4?\] is this your equation, think of the parent function, which in this case is \[\frac{ 1 }{ x^2 }\] it looks like |dw:1439265308847:dw| you can get this by simply making a table of values which you can do yourself. Once you have your parent function now we can do our transformations. \[y = \frac{ 3 }{ x^2 }\] where the 3 should give you a scaling factor then afterwards add the -4, \[y = \frac{ 3 }{ x^2 }-4\] where the -4 should give you a vertical translation of -4 units. If you're not comfortable at seeing it directly you can always make a table of values for each step, good luck!
anonymous
  • anonymous
No:( It is |dw:1439336098501:dw|
anonymous
  • anonymous
I'm sorry lol:)

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Astrophysics
  • Astrophysics
That means you're graph will look like this then |dw:1439336233138:dw|
Astrophysics
  • Astrophysics
Actually here, use this site to graph it out, but make sure you know how the transformations work https://www.desmos.com/calculator
anonymous
  • anonymous
Thanks:)! What about the domain, range
Astrophysics
  • Astrophysics
For domain solve \[x^2 - 4 \neq 0\] solve for x
anonymous
  • anonymous
um -2 or 2?
Astrophysics
  • Astrophysics
Yup, exactly
Astrophysics
  • Astrophysics
Notice that domain occupies the x values, and range is the y values so you have y>0 and \[y \le -3/4\]
anonymous
  • anonymous
ok:) And what about the VA? I have no idea what that means haha
Astrophysics
  • Astrophysics
vertical asymptotes
Astrophysics
  • Astrophysics
I'm assuming you don't know what limits are, so you see where it's not defined
Astrophysics
  • Astrophysics
So VA is at x = 2 and x = -2
anonymous
  • anonymous
And last one, I promise :) It says find the HA/SA???
Astrophysics
  • Astrophysics
I don't know what SA is but HA is horizontal asymptote
anonymous
  • anonymous
Okay so would that be -3/2?
Astrophysics
  • Astrophysics
Not quite, since the denominator has a higher degree than the numerator it's 0
Astrophysics
  • Astrophysics
That's just a shortcut though
Astrophysics
  • Astrophysics
You should actually look up what exactly this stuff means haha
anonymous
  • anonymous
lol! I probably should :) and I will
anonymous
  • anonymous
so just checking, so far we have: VA=-2,2 D-all real numbers r-y<=-3/4
Astrophysics
  • Astrophysics
domain is not all real numbers
Astrophysics
  • Astrophysics
X cannot be -2 and 2
anonymous
  • anonymous
oh.
Astrophysics
  • Astrophysics
\[D: x \in \mathbb{R}: x \neq \pm 2\]
Astrophysics
  • Astrophysics
It's all real except for + and - 2
anonymous
  • anonymous
ok:) Thanks :)
Astrophysics
  • Astrophysics
\[R: y \in \mathbb{R}: y \le - \frac{ 3 }{ 4 } ~or~y>0\]
anonymous
  • anonymous
So going back to the horizontal asym., I sort of get it know : http://www.purplemath.com/modules/asymtote2.htm
anonymous
  • anonymous
Thanks:)
anonymous
  • anonymous
so it would be 3/x^2?
Astrophysics
  • Astrophysics
y=0
anonymous
  • anonymous
Thanks so much:)))
Astrophysics
  • Astrophysics
and if the degree in the numerator was greater you'd have no horizontal asymptote, so like (x^3+2)/(x^2+3) no HA

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