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:
Solve for the first variable in one of the equations, then substitute the result into the other equation.
Point For:
(6111,4111,−411)(6111,4111,-411)
Equation Form:
x=6111,y=4111,z=−411x=6111,y=4111,z=-411
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=