11.1: Gráficas de funciones racionales
- Page ID
- 117683
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Recordemos desde el inicio de este capítulo que una función racional es una fracción de polinomios:
\[f(x)=\dfrac{a_n x^n+a_{n-1}x^{n-1}+\dots+a_1 x+a_0}{b_m x^m+b_{m-1}x^{m-1}+\dots+b_1 x+b_0} \nonumber \]
En esta sección, estudiaremos algunas características de gráficas de funciones racionales. Las asíntotas son características importantes de las gráficas de funciones racionales. Una asíntota es una línea que es abordada por la gráfica de una función:\(x=a\) es una asíntota vertical si se\(f(x)\) aproxima\(\pm \infty\) como\(x\) aproximaciones ya sea\(a\) desde la izquierda o desde la derecha, y\(y=b\) es una asíntota horizontal si\(f(x)\) se aproxima\(b\) como\(x\) enfoques\(\infty\) o\(-\infty\).
Para tener una mejor idea de algunas de estas características, incluidas las asíntotas, comenzamos por graficar algunas funciones racionales. En particular es útil conocer las gráficas de las funciones básicas\(y=\dfrac 1 {x^n}\).
Graficando\(y=\dfrac 1 {x}\),\(y=\dfrac 1 {x^2}\)\(y=\dfrac 1 {x^3}\),\(y=\dfrac 1 {x^4}\),, obtenemos:
En general vemos que\(x=0\) es una asíntota vertical y\(y=0\) es una asíntota horizontal. La forma de\(y=\dfrac 1 {x^n}\) depende de\(n\) ser par o impar. Contamos con:
Ahora estudiamos funciones racionales más generales.
- Nuestra primera gráfica es\(f(x)=\dfrac 1 {x-3}\).
Aquí, el dominio son todos los números donde el denominador no es cero, es decir\(D=\mathbb{R}-\{3\}\). Hay una asíntota vertical,\(x=3\). Además, la gráfica se aproxima\(0\) como\(x\) enfoques\(\pm \infty\). Por lo tanto,\(f\) tiene una asíntota horizontal,\(y=0\). En efecto, siempre que el denominador tenga un grado mayor que el numerador, la línea\(y=0\) será la asíntota horizontal.
- A continuación, graficamos\(f(x)=\dfrac{8x^2-8}{4x^2-16}\).
Aquí, el dominio es todo\(x\) para el cual\(4x^2-16\neq 0\). Para ver dónde sucede esto, calcule
\[4x^2-16=0 \implies \quad 4x^2=16 \implies\quad x^2=4 \implies\quad x=\pm 2 \nonumber \]
Por lo tanto, el dominio es\(D=\mathbb{R}-\{-2,2\}\). Como antes, vemos en la gráfica, que el dominio revela las asíntotas verticales\(x=2\) y\(x=-2\) (las líneas discontinuas verticales). Para encontrar la asíntota horizontal (la línea discontinua horizontal), observamos que cuando\(x\) se vuelve muy grande, los términos más altos tanto del numerador como del denominador dominan el valor de la función, de manera que
\[\text{for $|x|$ very large } \implies \quad f(x)=\dfrac{8x^2-8}{4x^2-16}\approx \dfrac{8x^2}{4x^2}=2 \nonumber \]
Por lo tanto, cuando se\(x\) acerca\(\pm\infty\), el valor de la función\(f(x)\) se acerca\(2\), y por lo tanto la asíntota horizontal está en\(y=2\) (la línea discontinua horizontal).
- Nuestra siguiente gráfica es\(f(x)=\dfrac{x^2-8x+15}{x-3}\).
Vemos que no parece haber ninguna asíntota vertical, a pesar de que no\(3\) está en el dominio. La razón de esto es que podemos “eliminar la singularidad” cancelando el término preocupante de la\(x-3\) siguiente manera:
\[f(x)=\dfrac{x^2-8x+15}{x-3}=\dfrac{(x-3)(x-5)}{(x-3)}=\dfrac{x-5}{1}=x-5,\quad x\not=3 \nonumber \]
Por lo tanto, la función se\(f\) reduce a\(x-5\) para todos los valores donde se define. Sin embargo, tenga en cuenta que no\(f(x)=\dfrac{x^2-8x+15}{x-3}\) se define en\(x=3\). Denotamos esto en la gráfica por un círculo abierto en\(x=3\), y lo llamamos una singularidad removible (o un agujero).
- Nuestra cuarta y última gráfica antes de exponer las reglas en plena generalidad es\(f(x)=\dfrac{2x^3-8}{3x^2-16}\).
La gráfica indica que no hay asíntota horizontal, ya que la gráfica parece aumentar hacia\(\infty\) y disminuir hacia\(-\infty\). Para hacer esta observación precisa, calculamos el comportamiento cuando se\(x\) aproxima\(\pm \infty\) ignorando los términos inferiores en el numerador y denominador
\[\text{for $|x|$ very large } \implies \quad f(x)=\dfrac{2x^3-8}{3x^2-16}\approx \dfrac{2x^3}{3x^2}=\dfrac{2x}{3} \nonumber \]
Por lo tanto, cuando\(x\) se vuelve muy grande, se\(f(x)\) comporta como\(\dfrac{2}{3}x\), que\(\infty\) se acerca cuando se\(x\) acerca\(\infty\), y\(-\infty\) se acerca cuando se\(x\) acerca \(-\infty\). (De hecho, después de realizar una división larga obtenemos\(\dfrac{2x^3-8}{3x^2-16}=\dfrac{2}{3}\cdot x+\dfrac{r(x)}{3x^2-16}\), lo que daría lugar a lo que se denomina asíntota inclinada\(y=\dfrac{2}{3}\cdot x\); véase también Nota: Comportamiento asintótico a continuación.) En efecto, siempre que el grado del numerador es mayor que el grado del denominador, encontramos que no hay asíntota horizontal, sino que la gráfica sopla hasta\(\pm\infty\). (Compárelo también con el ejemplo (c) anterior).
Combinando lo que hemos aprendido de los ejemplos anteriores, exponemos nuestras observaciones de la siguiente manera.
Dejar\(f(x)=\dfrac{p(x)}{q(x)}\) ser una función racional con polinomios\(p(x)\) y\(q(x)\) en el numerador y denominador, respectivamente.
- El dominio de\(f\) son todos los números reales\(x\) para los que el denominador no es cero,\[D=\{\quad x\in \mathbb{R} \quad|\quad q(x)\neq 0\quad \} \nonumber \]
- Supongamos que\(q(x_0)=0\), por lo que no\(f\) se define en\(x_0\). Si no\(x_0\) es una raíz de\(p(x)\), o si\(x_0\) es una raíz de\(p(x)\) pero de una multiplicidad menor que la raíz adentro\(q(x)\), entonces\(f\) tiene una asíntota vertical \(x=x_0\).
- Si\(p(x_0)=0\) y\(q(x_0)=0\), y la multiplicidad de la raíz\(x_0\) adentro\(p(x)\) es al menos la multiplicidad de la raíz adentro\(q(x)\), entonces estas raíces pueden ser canceladas, y se dice que hay un removible discontinuidad (o a veces llamado un agujero) en\(x=x_0\).
- Para encontrar las asíntotas horizontales, necesitamos distinguir los casos donde el grado de\(p(x)\) es menor que, igual o mayor que\(q(x)\).
Además, a veces es útil determinar las\(x\) - y\(y\) -intercepciones.
- Si\(0\) está en el dominio de\(f\), entonces la \(y\)-intercepción es\((0,f(0))\). \[f(x)=\dfrac{x+2}{x+1} \nonumber \]
- Si\(p(x_0)=0\) but \(q(x_0)\neq 0\), then \(f(x_0)=\dfrac{p(x_0)}{q(x_0)}=\dfrac{0}{q(x_0)}=0\), so that \((x_0,0)\) is an \(x\)-intercept, that is the graph intersects with the \(x\)-axis at \(x_0\). &&
We can calculate the asymptotic behavior of a rational function \(f(x)=\dfrac{p(x)}{q(x)}\) in the case where the degree of \(p\) is greater than the degree of \(q\) by performing a long division. If the the quotient is \(m(x)\), and the remainder is \(r(x)\), then
\[f(x)=\dfrac{p(x)}{q(x)}=m(x)+\dfrac{r(x)}{q(x)} \nonumber \]
Now, since deg\((r)<\)deg\((q)\), the fraction \(\dfrac{r(x)}{q(x)}\) approaches zero as \(x\) approaches \(\pm\infty\), so that \(f(x)\approx m(x)\) for large \(x\).
Find the domain, all horizontal asymptotes, vertical asymptotes, removable singularities, and \(x\)- and \(y\)-intercepts. Use this information together with the graph of the calculator to sketch the graph of \(f\).
- \(f(x)=\dfrac{-x^2}{x^2-3x-4}\)
- \(f(x)=\dfrac{5x}{x^2-2x}\)
- \(f(x)=\dfrac{x^3-9x^2+26x-24}{x^2-x-2}\)
- \(f(x)=\dfrac{x-4}{(x-2)^2}\)
- \(f(x)=\dfrac{3x^2-12}{2x^2+1}\)
Solution
- We combine our knowledge of rational functions and its algebra with the particular graph of the function. The calculator gives the following graph.
To find the domain of \(f\) we only need to exclude from the real numbers those \(x\) that make the denominator zero. Since \(x^2-3x-4=0\) exactly when \((x+1)(x-4)=0\), which gives \(x=-1\) or \(x=4\), we have the domain:
\[\text{domain }D=\mathbb{R}-\{-1,4\} \nonumber \]
The numerator has a root exactly when \(-x^2=0\), that is \(x=0\). Therefore, \(x=-1\) and \(x=4\) are vertical asymptotes, and since we cannot cancel terms in the fraction, there is no removable singularity. Furthermore, since \(f(x)=0\) exactly when the numerator is zero, the only \(x\)-intercept is \((0,0)\).
To find the horizontal asymptote, we consider \(f(x)\) for large values of \(x\) by ignoring the lower order terms in numerator and denominator,
\[|x| \text{ large } \implies f(x)\approx \dfrac{-x^2}{x^2}=-1 \nonumber \]
We see that the horizontal asymptote is \(y=-1\). Finally, for the \(y\)-intercept, we calculate \(f(0)\):
\[f(0)=\dfrac{- 0^2}{0^2-3\cdot 0-4}=\dfrac{0}{-4}=0 \nonumber \]
Therefore, the \(y\)-intercept is \((0,0)\). The function is then graphed as follows.
- The graph of \(f(x)=\dfrac{5x}{x^2-2x}\) as drawn with the TI-84 is the following.
For the domain, we find the roots of the denominator,
\[x^2-2x=0 \implies \quad x (x-2)=0 \implies \quad x=0 \text{ or } x=2 \nonumber \]
The domain is \(D=\mathbb{R}-\{0,2\}\). For the vertical asymptotes and removable singularities, we calculate the roots of the numerator,
\[5x=0 \implies \quad x=0 \nonumber \]
Therefore, \(x=2\) is a vertical asymptote, and \(x=0\) is a removable singularity. Furthermore, the denominator has a higher degree than the numerator, so that \(y=0\) is the horizontal asymptote. For the \(y\)-intercept, we calculate \(f(0)\) by evaluating the fraction \(f(x)\) at \(0\)
\[\dfrac{5\cdot 0}{0^2-2\cdot 0}=\dfrac{0}{0} \nonumber \]
which is undefined. Therefore, there is no \(y\)-intercept (we, of course, already noted that there is a removable singularity when \(x=0\)). Finally for the \(x\)-intercept, we need to analyze where \(f(x)=0\), that is where \(5x=0\). The only candidate is \(x=0\) for which \(f\) is undefined. Again, we see that there is no \(x\)-intercept. The function is then graphed as follows. (Notice in particular the removable singularity at \(x=0\).)
- We start again by graphing the function \(f(x)=\dfrac{x^3-9x^2+26x-24}{x^2-x-2}\) with the calculator. After zooming to an appropriate window, we get:
To find the domain of \(f\), we find the zeros of the denominator
\[x^2-x-2=0 \implies\quad (x+1)(x-2)=0\implies \quad x=-1\text{ or }x=2 \nonumber \]
The domain is \(D=\mathbb{R}-\{-1,2\}\). The graph suggests that there is a vertical asymptote \(x=-1\). However the \(x=2\) appears not to be a vertical asymptote. This would happen when \(x=2\) is a removable singularity, that is, \(x=2\) is a root of both numerator and denominator of \(f(x)=\dfrac{p(x)}{q(x)}\). To confirm this, we calculate the numerator \(p(x)\) at \(x=2\):
\[p(2)=2^3-9\cdot 2^2+26 \cdot 2-24=8-36+52-24=0 \nonumber \]
Therefore, \(x=2\) is indeed a removable singularity. To analyze \(f\) further, we also factor the numerator. Using the factor theorem, we know that \(x-2\) is a factor of the numerator. Its quotient is calculated via long division.
With this, we obtain
\[f(x)=\dfrac{(x-2)(x^2-7x+12)}{x^2-x-2}=\dfrac{(x-2)(x-3)(x-4)}{(x+1)(x-2)} \nonumber \]
Therefore, we conclude that \(x=-1\) is a vertical asymptote and \(x=2\) is a removable singularity. We also see that the \(x\)-intercepts are \((3,0)\) and \((4,0)\) (that is \(x-\) values where the numerator is zero).
Now, the long range behavior is determined by ignoring the lower terms in the fraction,
\[|x| \text{ large } \implies \quad f(x)\approx \dfrac{x^3}{x^2}=x \implies \quad \text{ no horizontal asymptote} \nonumber \]
Finally, the \(y\)-coordinate of the \(y\)-intercept is given by
\[y=f(0)=\dfrac{0^3-9\cdot 0^2+26\cdot 0-24}{0^2-0-2}=\dfrac{-24}{-2}=12 \nonumber \]
We draw the graph as follows:
- We first graph \(f(x)=\dfrac{x-4}{(x-2)^2}\). (Don’t forget to set the viewing window back to the standard settings!)
The domain is all real numbers except where the denominator becomes zero, that is, \(D=\mathbb{R}-\{2\}\). The graph has a vertical asymptote \(x=2\) and no hole. The horizontal asymptote is at \(y=0\), since the denominator has a higher degree than the numerator. The \(y\)-intercept is at \(y_0=f(0)=\dfrac{0-4}{(0-2)^2}=\dfrac{-4}{4}=-1\). The \(x\)-intercept is where the numerator is zero, \(x-4=0\), that is at \(x=4\). Since the above graph did not show the \(x\)-intercept, we can confirm this by changing the window size as follows:
Note in particular that the graph intersects the \(x\)-axis at \(x=4\) and then changes its direction to approach the \(x\)-axis from above. A graph of the function \(f\) which includes all these features is displayed below.
- We graph \(f(x)=\dfrac{3x^2-12}{2x^2+1}\).
For the domain, we determine the zeros of the denominator.
\[2x^2+1=0\implies \quad 2x^2=-1\implies \quad x^2=-\dfrac{1}{2} \nonumber \]
The only solutions of this equation are given by complex numbers, but not by any real numbers. In particular, for any real number \(x\), the denominator of \(f(x)\) is not zero. The domain of \(f\) is all real numbers, \(D=\mathbb{R}\). This implies in turn that there are no vertical asymptotes, and no removable singularities.
The \(x\)-intercepts are determined by \(f(x)=0\), that is where the numerator is zero,
\[3x^2-12=0\implies \quad 3x^2=12\implies \quad x^2=4\implies \quad x=\pm 2 \nonumber \]
The horizontal asymptote is given by \(f(x)\approx\dfrac{3x^2}{2x^2}=\dfrac 3 2\), that is, it is at \(y=\dfrac 3 2=1.5\). The \(y\)-intercept is at
\[y=f(0)=\dfrac{3\cdot 0^2-12}{2\cdot 0^2+1}=\dfrac{-12}{1}=-12 \nonumber \]
We sketch the graph as follows:
Since the graph is symmetric with respect to the \(y\)-axis, we can make one more observation, namely that the function \(f\) is even (see Observation: Even-Odd on page ):
\[f(-x)=\dfrac{3(-x)^2-12}{2(-x)^2+1}=\dfrac{3x^2-12}{2x^2+1}=f(x) \nonumber \]