Objective 2 - Graph Rational functions

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Convert between a rational function and its graph.

Link to section in online textbook.

You can print out these notes to follow along with the video below and keep notes to organize your thoughts.

We will work with two specific rational functions: $f(x) = \frac {1}{x}$ and $g(x)=\frac {1}{x^2}$ . By using what we know about shifting and leading coefficients from Quadratics, Radicals, and Polynomials, we have two basic equations for rational functions:

$f(x) = \frac {a}{x-h} + k$

$\graph [rectangular]{f(x) = a/(x-h) + k, a=1, h=0, k=0, x=h, y=k}$

$g(x) = \frac {a}{(x-h)^2} + k$

$\graph [rectangular]{f(x) = a/(x-h)^2 + k, a=1, h=0, k=0, x=h, y=k}$

Thinking back to the previous objective, our rational functions are not defined at $x=h$ . So while $(h, k)$ acts like our vertex for quadratics, it is not actually a point on the graph! Check out the Desmos graphs to see how $a$ , $h$ , and $k$ affect the graphs of these two functions.

We will focus on working from graphs to the equation. If you master this, you’ll be able to work backwards and graph a radical function from the equation.

Write an equation of the function graphed below. Assume $a = 1$ or $a=-1$ .

$f(x) = \frac {\answer {1}}{(x - \answer {0})^{\answer {1}}} + \answer {-2}$

The leading coefficient is either $a=1$ or $a=-1$ . Try going back to the Desmos graphs and switch between 1 and -1. For the other parts, what acts like the “vertex” of the graph?