Answer
A. The domain is $(-\infty,-5) \cup(-5,5)\cup (5, \infty)$
B. The y-intercept is $0$
The x-intercept is $0$
C. The function is not an odd function or an even function.
D. $y = -1$ is a horizontal asymptote.
$x = 5$ is a vertical asymptote.
$x = -5$ is a point discontinuity.
E. The function is increasing on the intervals $(-\infty, -5)\cup (-5,5)\cup (5,\infty)$
F. There is no local maximum or local minimum.
G. The graph is concave up on the interval $(-\infty, -5)\cup (-5,5)$
The graph is concave down on the interval $(5, \infty)$
There no points of inflection.
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{x^2+5x}{25-x^2}$
A. The function is defined for all real numbers except $x = -5$ and $x = 5$.
The domain is $(-\infty,-5) \cup(-5,5)\cup (5, \infty)$
B. When $x = 0,$ then $y = \frac{(0)^2+5(0)}{25-(0)^2}= 0$
The y-intercept is $0$
When $y = 0$:
$\frac{x^2+5x}{25-x^2} = 0$
$x (x+5)= 0$
$x = 0, -5$
Note that the function is undefined when $x = -5$
The x-intercept is $0$
C. The function is not an odd function or an even function.
D. $\lim\limits_{x \to -\infty} (\frac{x^2+5x}{25-x^2}) = -1$
$\lim\limits_{x \to \infty} (\frac{x^2+5x}{25-x^2}) = -1$
$y = -1$ is a horizontal asymptote.
$\lim\limits_{x \to 5^-} (\frac{x^2+5x}{25-x^2}) = \infty$
$\lim\limits_{x \to 5^+} (\frac{x^2+5x}{25-x^2}) = -\infty$
$x = 5$ is a vertical asymptote.
$\lim\limits_{x \to -5^-} (\frac{x^2+5x}{25-x^2}) = -\frac{1}{2}$
$\lim\limits_{x \to -5^+} (\frac{x^2+5x}{25-x^2}) = -\frac{1}{2}$
$x = -5$ is a point discontinuity.
E. We can try to find the values of $x$ such that $y' = 0$:
$y' = \frac{(2x+5)(25-x^2)-(x^2+5x)(-2x)}{(25-x^2)^2}$
$y' = \frac{5x^2+50x+125}{(5-x)^2(5+x)^2}$
$y' = \frac{5(x^2+10x+25)}{(5-x)^2(5+x)^2}$
$y' = \frac{5(x+5)^2}{(5-x)^2(5+x)^2}$
$y' = \frac{5}{(5-x)^2}$
There are no values of $x$ such that $y'=0$
When $x \lt -5$ or $-5 \lt x \lt 5$ or $x \gt 5$, then $y' \gt 0$
The function is increasing on the intervals $(-\infty, -5)\cup (-5,5)\cup (5,\infty)$
F. There is no local maximum or local minimum.
G. We can try to find the values of $x$ such that $y'' = 0$:
$y'' = \frac{-5(2)(5-x)(-1)}{(5-x)^4}$
$y'' = \frac{10}{(5-x)^3}$
There are no values of $x$ such that $y''=0$
When $x \lt -5~~$ or $-5 \lt x \lt 5$, then $y'' \gt 0$
The graph is concave up on the interval $(-\infty, -5)\cup (-5,5)$
When $x \gt 5$, then $y'' \lt 0$
The graph is concave down on the interval $(5, \infty)$
There no points of inflection.
H. We can see a sketch of the curve below.
