$f\left(x\right)=\frac{1}{x}$ is stretched vertically by a scale factor of $k$.
Find the equation of the new function when $k=2$.$\qquad p\left(x\right)=$
Find the equation of the new function when $k=3$.$\qquad q\left(x\right)=$
Find the equation of the new function when $k=4$.$\qquad r\left(x\right)=$
Write down the equation of the vertical asymptote of $f\left(x\right)=\frac{k}{x}$ where $k\ne0$ is a constant.
Write down the equation of the horizontal asymptote of $f\left(x\right)=\frac{k}{x}$ where $k\ne0$ is a constant.
The domain of $f\left(x\right)=\frac{k}{x}$ where $k \ne 0$ is a constant is $x\ne$
$, x\in$
.
The range of $f\left(x\right)=\frac{k}{x}$ where $k \ne 0$ is a constant is $y\ne$
$, y\in$
.
Reflecting $f\left(x\right)=\frac{1}{x}$ in the $x$-axis
$f\left(x\right)=\frac{1}{x}$ is reflected in the $x$-axis.
Find the equation of the new function.$\qquad h\left(x\right)=$
Write down the equation of the vertical asymptote of $h\left(x\right)$.
Write down the equation of the horizontal asymptote of $h\left(x\right)$.
The domain of $h\left(x\right)=-\frac{1}{x}$ is $x\ne$
$, x\in$
.
The range of $h\left(x\right)=-\frac{1}{x}$ is $y\ne$
$, y\in$
.
Translating $f\left(x\right)=\frac{1}{x}$
$f\left(x\right)=\frac{1}{x}$ is translated by the vector $\binom{-2}{3}$.
Find the equation of the new function.$\qquad g\left(x\right)=$
Write down the equation of the vertical asymptote of $g\left(x\right)$.
Write down the equation of the horizontal asymptote of $g\left(x\right)$.
The domain of $g\left(x\right)=\frac{1}{x+2}+3$ is $x\ne$
$, x\in$
.
The range of $g\left(x\right)=\frac{1}{x+2}+3$ is $y\ne$
$, y\in$
.
Transforming $f\left(x\right)=\frac{1}{x}$
$f\left(x\right)=\frac{1}{x}$ is stretched vertically by scale factor 2, then reflected in the $x$-axis, then translated by the vector $\binom{-2}{3}$.
Find the equation of the new function.$\qquad t\left(x\right)=$
Write down the equation of the vertical asymptote of $t\left(x\right)$.
Write down the equation of the horizontal asymptote of $t\left(x\right)$.
The domain of $t\left(x\right)=-\frac{2}{x+2}+3$ is $x\ne$
$, x\in$
.
The range of $t\left(x\right)=-\frac{2}{x+2}+3$ is $y\ne$
$, y\in$
.
Practice
Find the equation of the horizontal asymptote for each of the following functions.
$f\left(x\right)=\frac{1}{x-3}+4$
$g\left(x\right)=\frac{3}{x}-2$
$h\left(x\right)=-\frac{2}{x+5}$
$j\left(x\right)=-\frac{1}{x-2}+6$
$k\left(x\right)=4+\frac{1}{x-3}$
$l\left(x\right)=6-\frac{1}{x-2}$
Practice
Find the equation of the vertical asymptote for each of the following functions.