The question states:
A). Find the vertical asymptotes of the functions graph.
B). Describe the functions end behavior using infinity notation.
G(x)= 4x^2/12x^2+x-1
Please give answer and show work so I can understand.
I could not get A). But for B). I got as x>+infinity, g(x)> 1/3 and as x>-infinity, g(x)>1/3
A). Find the vertical asymptotes of the functions graph.
B). Describe the functions end behavior using infinity notation.
G(x)= 4x^2/12x^2+x-1
Please give answer and show work so I can understand.
I could not get A). But for B). I got as x>+infinity, g(x)> 1/3 and as x>-infinity, g(x)>1/3
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I am a bit confused by your question. The function you have given is:
G(x) = (4x^2)/(12x^2) + x - 1.
If this is in fact the function you are asking about, then the question is mistaken. It should be obvious that the first term, (4x^2)/(12x^2) can be reduced to 1/3 provided x does not equal 0. This leaves us with a simple linear equation: G(x) = 1/3 + x - 1, for x not equal to 0.
Now, in the case that x does equal 0, the function does not provide an asymptote, but rather a removable discontinuity at (0,-2/3), which we usually draw as an empty hole in the function in calculus.
As stated, this function G(x) = (4x^2)/(12x^2) + x - 1 has no asymptotes, either vertical or horizontal. It is a simple linear function everywhere except 0, so as x > + infinity, G(x) > + infinity and x > - infinity, G(x) > - infinity, which is true for any linear function that has a positive slope.
Ohh, now I see. Reading over your question again, I realize you meant to say:
G(x) = 4x^2/(12x^2 + x - 1).
This makes things very different.
Okay so the vertical asymptotes for this function are when the function's denominator equals 0. So we set 12x^2 + x - 1 = 0 and solve for x. We can factor this into (4x - 1)(3x + 1) = 0. So there are vertical asymptotes at x = 1/4 and x = -1/3. In neither case does the numerator also equal 0 when the denominator equals 0 in this problem, so they both represent vertical asymptotes rather than removable discontinuities.
As for the horizontal asymptotes (also known as end behavior using infinity notation):
x > + infinity, G(x) > 1/3 and x > - infinity, G(x) > 1/3, which can be seen by observing that as x becomes very large the values of G(x) tend to go toward 1/3. If you have a calculator, you could just input G(10000) or some similiarly large number and get a number very close to 1/3, which is convincing evidence. The same can be said of x becoming very negative, as x goes to - infinity, the values tend toward 1/3. Once again G(-10000) is very close to 1/3.
G(x) = (4x^2)/(12x^2) + x - 1.
If this is in fact the function you are asking about, then the question is mistaken. It should be obvious that the first term, (4x^2)/(12x^2) can be reduced to 1/3 provided x does not equal 0. This leaves us with a simple linear equation: G(x) = 1/3 + x - 1, for x not equal to 0.
Now, in the case that x does equal 0, the function does not provide an asymptote, but rather a removable discontinuity at (0,-2/3), which we usually draw as an empty hole in the function in calculus.
As stated, this function G(x) = (4x^2)/(12x^2) + x - 1 has no asymptotes, either vertical or horizontal. It is a simple linear function everywhere except 0, so as x > + infinity, G(x) > + infinity and x > - infinity, G(x) > - infinity, which is true for any linear function that has a positive slope.
Ohh, now I see. Reading over your question again, I realize you meant to say:
G(x) = 4x^2/(12x^2 + x - 1).
This makes things very different.
Okay so the vertical asymptotes for this function are when the function's denominator equals 0. So we set 12x^2 + x - 1 = 0 and solve for x. We can factor this into (4x - 1)(3x + 1) = 0. So there are vertical asymptotes at x = 1/4 and x = -1/3. In neither case does the numerator also equal 0 when the denominator equals 0 in this problem, so they both represent vertical asymptotes rather than removable discontinuities.
As for the horizontal asymptotes (also known as end behavior using infinity notation):
x > + infinity, G(x) > 1/3 and x > - infinity, G(x) > 1/3, which can be seen by observing that as x becomes very large the values of G(x) tend to go toward 1/3. If you have a calculator, you could just input G(10000) or some similiarly large number and get a number very close to 1/3, which is convincing evidence. The same can be said of x becoming very negative, as x goes to - infinity, the values tend toward 1/3. Once again G(-10000) is very close to 1/3.
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