# asymptotes and intercepts

#### vaironxxrd

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 = $$\displaystyle \sqrt{3}$$

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#### skeeter

MHF Helper
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
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?

#### vaironxxrd

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

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?

#### vaironxxrd

... 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
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.

#### vaironxxrd

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
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 $$\displaystyle x = -\sqrt{3}$$