This shows that the function is undefined at
$y=q$ . Therefore the range of
$f\left(x\right)=\frac{a}{x}+q$ is
$\left\{f\right(x):f(x)\in (-\infty ;q)\cup (q;\infty \left)\right\}$ .
For example, the domain of
$g\left(x\right)=\frac{2}{x}+2$ is
$\{x:x\in \mathbb{R},x\ne 0\}$ because
$g\left(x\right)$ is undefined at
$x=0$ .
We see that
$g\left(x\right)$ is undefined at
$y=2$ . Therefore the range is
$\left\{g\right(x):g(x)\in (-\infty ;2)\cup (2;\infty \left)\right\}$ .
Intercepts
For functions of the form,
$y=\frac{a}{x}+q$ , the intercepts with the
$x$ and
$y$ axis is calculated by setting
$x=0$ for the
$y$ -intercept and by setting
$y=0$ for the
$x$ -intercept.
There are two asymptotes for functions of the form
$y=\frac{a}{x}+q$ . Just a reminder, an asymptote is a straight or curved line, which the graph of a function will approach, but never touch. They are determined by examining the domain and range.
We saw that the function was undefined at
$x=0$ and for
$y=q$ . Therefore the asymptotes are
$x=0$ and
$y=q$ .
For example, the domain of
$g\left(x\right)=\frac{2}{x}+2$ is
$\{x:x\in \mathbb{R},x\ne 0\}$ because
$g\left(x\right)$ is undefined at
$x=0$ . We also see that
$g\left(x\right)$ is undefined at
$y=2$ . Therefore the range is
$\left\{g\right(x):g(x)\in (-\infty ;2)\cup (2;\infty \left)\right\}$ .
From this we deduce that the asymptotes are at
$x=0$ and
$y=2$ .
Sketching graphs of the form
$f\left(x\right)=\frac{a}{x}+q$
In order to sketch graphs of functions of the form,
$f\left(x\right)=\frac{a}{x}+q$ , we need to determine four characteristics:
domain and range
asymptotes
$y$ -intercept
$x$ -intercept
For example, sketch the graph of
$g\left(x\right)=\frac{2}{x}+2$ . Mark the intercepts and asymptotes.
We have determined the domain to be
$\{x:x\in \mathbb{R},x\ne 0\}$ and the range to be
$\left\{g\right(x):g(x)\in (-\infty ;2)\cup (2;\infty \left)\right\}$ . Therefore the asymptotes are at
$x=0$ and
$y=2$ .
There is no
$y$ -intercept and the
$x$ -intercept is
${x}_{int}=-1$ .
Draw the graph of
$y=\frac{-4}{x}+7$ .
The domain is:
$\{x:x\in \mathbb{R},x\ne 0\}$ and the range is:
$\left\{f\right(x):f(x)\in (-\infty ;7)\cup (7;\infty \left)\right\}$ .
We look at the domain and range to determine where the asymptotes lie. From the domain we see that the function is undefined when
$x=0$ , so there is one asymptote at
$x=0$ . The other asymptote is found from the range. The function is undefined at
$y=q$ and so the second asymptote is at
$y=7$
There is no y-intercept for graphs of this form.
The x-intercept occurs when
$y=0$ . Calculating the x-intercept gives:
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