Find horizontal asymptote(s) and vertical asymptote(s) of..
- Thread starter Jade
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What you have posted means the following:
. . . . .\(\displaystyle \L f(x)\, =\, -3x^2\, + \,2x\, - \,\frac{1}{x^2}\, +\, 2x\, - \, 15\)
Is that what you meant? Or did you mean this?
. . . . .\(\displaystyle \L f(x)\, =\, \frac{-3x^2\, + \, 2x\, - \, 1}{x^2\, +\, 2x\, - \, 15}\)
Or something else...?
Finding the vertical and horizontal asymptotes is fairly formulaic and was covered back in algebra. Where are you stuck?
Please be specific. Thank you.
Eliz.
. . . . .\(\displaystyle \L f(x)\, =\, -3x^2\, + \,2x\, - \,\frac{1}{x^2}\, +\, 2x\, - \, 15\)
Is that what you meant? Or did you mean this?
. . . . .\(\displaystyle \L f(x)\, =\, \frac{-3x^2\, + \, 2x\, - \, 1}{x^2\, +\, 2x\, - \, 15}\)
Or something else...?
Finding the vertical and horizontal asymptotes is fairly formulaic and was covered back in algebra. Where are you stuck?
Please be specific. Thank you.
Eliz.
\(\displaystyle \L\\f(x)=\frac{-3x^{2}+2x-1}{x^{2}+2x-15}\)
To find the vertical asymptotes, find what makes the denominator equal to 0.
For the horizontal asymptote, what does f(x) approach as x approaches +/- infinity?.
Another observation: Since the power of the numerator and the denominator are the same, the ratio of the leading coefficients is a horizontal asymptote.
To find the vertical asymptotes, find what makes the denominator equal to 0.
For the horizontal asymptote, what does f(x) approach as x approaches +/- infinity?.
Another observation: Since the power of the numerator and the denominator are the same, the ratio of the leading coefficients is a horizontal asymptote.
