for the graph:
such that the function is : $ y = \frac{a+x}{b+cx} $ where a = -2, b = 1 and c = 1/2
how do you sketch the graph of $ y = |\frac{b+cx}{a+x}| $ ??
i got that the VA of the new graph is "$+2$" and the root of the new graph is "$-2$", but then the answer says that the horizontal asymptote is "$\frac{1}{2}$" which i don't really understand..
Could anyone explain to me what to happen, if i am to transform a graph from $f(x)$ to $\frac{1}{f(x)}$.
i only know that the roots of $f(x)$ will become VA value of $1/f(x)$ and VA values of $f(x)$ becomes the zeros of $\frac{1}{f(x)}$
and that if $f(x) \gt 0$ , then $\frac{1}{f(x)}$ $\gt 0$ such that $f(x) \to 0 \implies \frac{1}{f(x)} \to \infty$ and $f(x) \to \infty \implies \frac{1}{f(x)} \to 0$.
You have a function $$f(x) = \frac{x-2}{\frac{x}{2} + 1}$$ and you want to sketch $y = \frac{1}{f(x)}$.
The $y$-intercept of the graph $y = f(x)$ is given by $f(0)$ (let this be $y_0$). The $y$-intercept of the graph $y = \frac{1}{f(x)}$ is given by $\frac{1}{f(0)}$, which is the reciprocal of $y_0$.
Similarly, the old horizontal asymptote, $y = \displaystyle\lim_{x\to\infty}f(x) = 2$, will become $y = \displaystyle\lim_{x\to\infty} \frac{1}{f(x)} = \frac{1}{\displaystyle\lim_{x\to\infty}f(x)} = \frac{1}{2}$, since $f(x)$ is being replaced by $\frac{1}{f(x)}$.
In general, the $y$ values of $y = f(x)$ will be the reciprocals of $y = \frac{1}{f(x)}$, i.e.,
$$y_{\text{new}} = \frac{1}{y_{\text{old}}}.$$
To then sketch the absolute value of this function, simply reflect any part below the $x$-axis above it, so there are no negative $y$ values for any $x$.
Pictures speak a thousand words...
Note that for each $x$ value, the $y$ values are reciprocals:
Reflecting the part below the $x$-axis:
Also, technically $f$ is not defined at $x = -2$, so neither should $\frac{1}{f}$ or $\left | \frac{1}{f} \right |$, i.e., there are holes at $x=-2$ on the blue graphs.