# Find all vertical, horizontal, and slant asymptotes x- and y- intercepts, and

#### math951

symmetries, and then graph each function.

y=1/(x-2)

so the horizontal asymptote is y=1 because the denominator and numerator both = same degree.

Verical asymtpote is x=2 ( pretty easy.)

Slant asymtpote is none because? you cannot divide 1/x-2 right?

x intercept = I believe the x-intercept is none.

y-intercept = -1/2

How do I find the symmetries for this problem? And How do I graph it?

Thank you.

#### skeeter

MHF Helper
symmetries, and then graph each function.

y=1/(x-2)

so the horizontal asymptote is y=1 because the denominator and numerator both = same degree.

no ... horizontal asymptote is y = 0, review what "degree" means.

Verical asymtpote is x=2 ( pretty easy.)

Slant asymtpote is none because? you cannot divide 1/x-2 right?

not quite ... slant occurs when degree of numerator is one greater than degree of the denominator.

x intercept = I believe the x-intercept is none.

y-intercept = -1/2

How do I find the symmetries for this problem? And How do I graph it?

graph the parent function y = 1/x , then translate 2 units to the right.

#### math951

But for the horizontal asymptotes I thought if the degree of numerator is 1 and the degree of the deminator is 1, then you divide them?

#### skeeter

MHF Helper
$y = \dfrac{1}{x-2}$ ... degree of the numerator is 0 , degree of the denominator is 1

#### math951

$y = \dfrac{1}{x-2}$ ... degree of the numerator is 0 , degree of the denominator is 1
Can't 1 be the degree of any number technically? or is this because the 1 has no x.

So lets suppose the equation was : y=1x+1/ x-2 then y= 1 because they both have the same variable?

#### skeeter

MHF Helper
no ... the numerator is the constant $1 = x^0$, degree is zero.

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