document.write( "Question 116693: Graph this rational function finding the asymptotes,zeros,y-intercept, and other points as needed.\r
\n" ); document.write( "\n" ); document.write( " $\"f%28x%29=+%282x%5E2-8%29%2F%28x%5E2-25%29\"$ \n" ); document.write( "

Algebra.Com's Answer #84858 by solver91311(24713) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"f%28x%29=+%282x%5E2-8%29%2F%28x%5E2-25%29\"$ \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Vertical Asymptotes. These occur anywhere the function is undefined. Since the domains of the numerator and denominator functions are all reals, we only need to concern ourselves with values of x that would make the denominator zero and thus the entire rational function be undefined. Factoring the denominator gives us $\"%28x-5%29%28x%2B5%29\"$ , so we can see that the function is undefined at either 5 or -5. Therefore there are vertical asymptotes at $\"x=5\"$ and $\"x=-5\"$ .\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Horizontal Asymptotes. Rational functions have at most one horizontal asymptote. If the degree of the numerator polynomial is smaller than the degree of the denominator polynomial, then there is a horizontal asymptote at $\"y=0\"$ . If the degree of the numerator polynomial is larger than the degree of the denominator polynomial, then there is no horizontal asymptote at all. If the degree of the polynomials are equal, then there is a horizontal asymptote at the line $\"y=a%5B1%5D%2Fa%5B2%5D\"$ where $\"a%5B1%5D\"$ is the coefficient on the high order term in the numerator and $\"a%5B2%5D\"$ is the coefficient on the high order term in the denominator. In the case of the given function, the degrees are equal, so the horizontal asymptote is at $\"y=2%2F1=2\"$ .\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Zeros. These are the values of x that make the function equal zero. If you factor the numerator, you get $\"2%28x%5E2-4%29=2%28x-2%29%28x%2B2%29\"$ which has the numerator go to zero at 2 and -2. Note that when you do this in general, you should check to make sure that the zeros of the numerator polynomial are not also zeros of the denominator such that the entire rational expression is undefined. In that case, you would exclude those values for zeros of the function. In the given problem, that is not the case, so we can say the zeros are 2 and -2. In terms of ordered pairs, that would be (2, 0) and (-2, 0).\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "y-intercept. The y-intercept is the point on the y axis that represents the value of the function when x = 0, or $\"f%280%29=+%282%280%29%5E2-8%29%2F%280%5E2-25%29=%28-8%29%2F%28-25%29=8%2F25\"$ \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Other than that, you can pick some values on either side of the critical points--those where the graph is asymptotic, in the neighborhood of the zeros and the y-intercept, just to give yourself an idea of the general curvature of the function.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"graph%28600%2C600%2C-20%2C20%2C-20%2C20%2C%282x%5E2-8%29%2F%28x%5E2-25%29%29\"$ \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Ignore the slightly slanted straight lines on the either side of the graph. They only show up because this graphing system doesn't handle vertical asymptotes very well.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hope that helps,
\n" ); document.write( "John \n" ); document.write( "

\n" );