asymptotes and intercepts

Nov 2010
433
4
USA
I need to find the y,x intercepts and asymptotes of the following function

\(\displaystyle f(x)=\frac{2}{x^2-3}\)

x intercepts = No x intercents
y intercepts = \(\displaystyle \sqrt{3}\)
horizontal asymptote = 0
vertical asymptote = \(\displaystyle \sqrt{3}\)
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I need to find the y,x intercepts and asymptotes of the following function

\(\displaystyle f(x)=\frac{2}{3x-3}\)

x intercepts = No x intercents correct

y intercepts = \(\displaystyle \sqrt{3}\) no
horizontal asymptote = 0 y = 0 is the horizontal asymptote (it's the equation of a line)
vertical asymptote = \(\displaystyle \sqrt{3}\) no
... where did \(\displaystyle \sqrt{3}\) come from?
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
I need to find the y,x intercepts and asymptotes of the following function

\(\displaystyle f(x)=\frac{2}{3x-3}\)

x intercepts = No x intercents
y intercepts = \(\displaystyle \sqrt{3}\)
horizontal asymptote = 0
vertical asymptote = \(\displaystyle \sqrt{3}\)
y intercepts are where x = 0. This does not give \(\displaystyle \displaystyle \sqrt{3}\).

The vertical asymptote is at the same x value as the horizontal translation. Maybe try rewriting the function as \(\displaystyle \displaystyle \frac{2}{3}\left(\frac{1}{x - 1}\right) \). Does that help?
 
Nov 2010
433
4
USA
... where did \(\displaystyle \sqrt{3}\) come from?
y intercepts are where x = 0. This does not give \(\displaystyle \displaystyle \sqrt{3}\).

The vertical asymptote is at the same x value as the horizontal translation. Maybe try rewriting the function as \(\displaystyle \displaystyle \frac{2}{3}\left(\frac{1}{x - 1}\right) \). Does that help?
Sorry i made a big mistake, while looking at my paper i stated the numerator correctly but the denominator incorrectly, I have fixed it.
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
\(\displaystyle f(x) = \frac{2}{x^2-3}\)


the y-intercept is \(\displaystyle f(0)\) , which is not \(\displaystyle \sqrt{3}\)

note that the denominator will factor ... \(\displaystyle (x - \sqrt{3})(x + \sqrt{3})\)

now, what are the equations of the vertical asymptotes?
 
Nov 2010
433
4
USA
... where did \(\displaystyle \sqrt{3}\) come from?
y intercepts are where x = 0. This does not give \(\displaystyle \displaystyle \sqrt{3}\).

The vertical asymptote is at the same x value as the horizontal translation. Maybe try rewriting the function as \(\displaystyle \displaystyle \frac{2}{3}\left(\frac{1}{x - 1}\right) \). Does that help?
\(\displaystyle f(x) = \frac{2}{x^2-3}\)


the y-intercept is \(\displaystyle f(0)\) , which is not \(\displaystyle \sqrt{3}\)

note that the denominator will factor ... \(\displaystyle (x - \sqrt{3})(x + \sqrt{3})\)

now, what are the equations of the vertical asymptotes?
Let me see if I understand because my review paper is getting me a bit confused.
x and y intercepts are found by f(0)?
horizontals asymptotes are determined by the first degrees m & n
vertical asymptotes is found by setting the denomenator = 0?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
Let me see if I understand because my review paper is getting me a bit confused.
x and y intercepts are found by f(0)?
horizontals asymptotes are determined by the first degrees m & n
vertical asymptotes is found by setting the denomenator = 0?
f(0) is the y-intercept (only one)

values of x where f(x) = 0 are the x-intercepts (could be one, more than one ... or none at all)

horizontal asymptote is y = 0 when degree of numerator < degree of denominator, or y = ratio of the leading coefficients if degrees are equal.

vertical asymptotes are x = the value(s) where the function is undefined (and the discontinuities are non-removable) ... so yes, x-values where the denominator equals 0.
 
Nov 2010
433
4
USA
f(0) is the y-intercept (only one)

values of x where f(x) = 0 are the x-intercepts (could be one, more than one ... or none at all)

horizontal asymptote is y = 0 when degree of numerator < degree of denominator, or y = ratio of the leading coefficients if degrees are equal.

vertical asymptotes are x = the value(s) where the function is undefined (and the discontinuities are non-removable) ... so yes, x-values where the denominator equals 0.
In that case...
Y intercept = g(0)= (0)^2-3 = -3
vertical asymptote =\(\displaystyle x^2-3=0\)
\(\displaystyle x^2=3\)
\(\displaystyle x=\sqrt(3)\)
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
In that case...
Y intercept = g(0)= (0)^2-3 = -3
vertical asymptote =\(\displaystyle x^2-3=0\)
\(\displaystyle x^2=3\)
\(\displaystyle x=\sqrt(3)\)
\(\displaystyle f(0) = \frac{2}{0^2 - 3} = -\frac{2}{3}\)

vertical asymptotes are \(\displaystyle x = \sqrt{3}\) and \(\displaystyle x = -\sqrt{3}\)