Saturday, May 18, 2013

College Algebra, Chapter 4, Chapter Review, Section Review, Problem 72

Graph the rational function $\displaystyle r(x) = \frac{x^3 + 27}{x + 4}$. Show clearly all $x$ and $y$ intercepts and asymptotes.

We first find the factor $r$, so $\displaystyle r(x) = \frac{(x + 3)(x^2 - 3x+ 9)}{x + 4}$

$x$ intercept: The $x$ intercepts are the zeros of the numerator, $x = -3$

$y$ intercept: To find the $y$ intercept, we substitute $x = 0$.

$\displaystyle r(0) = \frac{0^3 + 27}{0 + 4} = \frac{27}{4}$

The $y$ intercept is $\displaystyle \frac{27}{4}$

Vertical Asymptotes: The vertical asymptotes occur where the denominator is , that is where the function is undefined. Hence, the line $x = -4$ is the vertical asymptote.

Next, to know the behavior near vertical asymptote, we set $-3.9$ and $-4.1$ from the right and left of $x = -4$ respectively. So,

$\displaystyle y = \frac{[(-3.9) + 3][(-3.9)^2 - 3(-3.9) + 9]}{[(-3.9) + 4]}$ whose sign is $\displaystyle \frac{(-)(+)}{(+)}$ (negative)

Hence, $y \to - \infty$ as $x \to -4^+$. Then by substituting $x = 4.1$ to the function, we will find out that $y \to \infty$ as $x \to -4^-$.

Since, the degree of the numerator is greater than the degree of the denominator, then the function has no horizontal asymptote but rather slant asymptote. So by applying long division







Thus, the polynomial function $y = x^2 - 4x + 16$ is the asymptote of the rational function.

Therefore, the graph will look like..

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