Graph 3/x^2-4. Then find the domain/range/VA

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Graph 3/x^2-4. Then find the domain/range/VA

Mathematics
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\[f(x) = \frac{ 3 }{ x^2 }-4?\] 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!
No:( It is |dw:1439336098501:dw|
I'm sorry lol:)

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