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