# Standard transformations can be used to help graph rational functions of the form f(x)=frac{A}{Bx-C}+D Explain how the parameters A, B, C, and D relate to the graph of the rational functions.

Standard transformations can be used to help graph rational functions of the form $f\left(x\right)=\frac{A}{Bx-C}+D$ Explain how the parameters A, B, C, and D relate to the graph of the rational functions.
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Step 1 Conclude function f(x) given: Has a vertical asymptote at $x=\frac{C}{B}$ Has a horizontal asymptote at $y=D$ Step 2 Additionally, we may conclude that: If $C\ne q0$, the function will have a y-intercept at $f\left(0\right)=-\frac{A}{C}+D$ If $D\ne q0$, we find the x-intercept as $\frac{A}{Bx-C}+D=0$ which we solve for x to obtain $x=\frac{CD-A}{BD}$ The function will have a domain of $\left(-\mathrm{\infty },\frac{C}{B}\right)\cup \left(\frac{C}{B},\mathrm{\infty }\right)$ If , the function will increase on its domain The end behavior of the function is $y\to D$ as y $\to ±\mathrm{\infty }$ Finally, the graph of f(x) is obtained from $g\left(x\right)=\frac{1}{x}$ by the following transformation: $\frac{1}{x}$ basic graph $\frac{1}{x-\frac{C}{B}}$ translate $\frac{C}{B}$ units to the right $\frac{A}{B}\cdot \frac{1}{x-\frac{C}{B}}$ vertically stretch by a factor of $\frac{A}{B}$ $\frac{A}{B}\cdot \frac{1}{x-\frac{C}{B}}+D$ translate D units up Simplifying the last exprexxion will result in the formula for f(x) aas given.