Identify the asymptote(s) of this function: y=5/(x-4)+1

1 answer:

6 0

$y= \frac{5}{(x-4)+1}$
$y= \frac{5}{x-3}$
So you take the denominator and compare to zero.
$x-4+1=0$
$x-3=0$
$x=3$
Then the horizontal asymptote is y = 0

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If point $\left(x,\ \frac{\sqrt{7}}{3}\right)$ is on the unit ircle, then:

$x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\ \\ \Rightarrow x^2+ \frac{7}{9} =1 \\ \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9} \\ \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}$

Since, the point is in the second quadrant, x is negative.

Thus, $(x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)$

Part A:

$3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\ \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\ \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\ \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta$

Therefore, $-6\tan\theta=3\sqrt{7}$ .

Part B:

$- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x} \\ \\ =\tan\theta$

Therefore, $\tan\theta=- \frac{ \sqrt{7} }{2}$