1) 2
x^2-2x-3
Find) Domain
Range
Is This Continuous?
Local min/max
Vertical Asymptote
Horizontal Asymptote
X Intercepts
Y Intercepts
for function $\displaystyle f(x) = \frac{2}{x^2-2x-3}$
$\displaystyle f(x) = \frac{2}{(x+1)(x-3)}$
Domain
Domain is the set of values of x for which function is defined.
The given function is not defined at x = -1 and x = 3
So, domain $\displaystyle = \{x \in \mathbb{R} | \;x\ne-1,3 \}$
Range
Range is f(x) values (or y values) you get, when you put these domain x values in function f(x). Since the function has a horizontal Asymptote at y = 0, so, y = 0 is excluded from Range.
So, Range $\displaystyle = \{y \in \mathbb{R} | \;y\ne0\}$
Horizontal Asymptote (HA)
To find HA, divide Numerator and denominator of the function with highest degree (power) of x. Then take limit $\displaystyle x \rightarrow \infty$
$\displaystyle f(x) = \frac{2}{x^2-2x-3}$
divide with highest degree, $\displaystyle x^2$,
$\displaystyle f(x) = \frac{\frac{2}{x^2}}{\frac{x^2}{x^2}-\frac{2x}{x^2}-\frac{3}{x^2}}$
$\displaystyle f(x) = \frac{\frac{2}{x^2}}{1-\frac{2}{x}-\frac{3}{x^2}}$
Now take $\displaystyle \lim{x\to \infty}$
$\displaystyle = \frac{0}{1-0-0}=0$
HA is y = 0
Vertical Asymptote (VA)
For VA, Denominator = 0
(x+1)(x-3) = 0
x = -1 and x = 3
The graph is discontinuous at x = -1 and x = 3
y-intercepts
For y-intercept, put, x = 0 in equation,
we get, y = -2/3
y-intercept = -2/3, the graph cuts y-axis at -2/3.
x-intercepts
For x-intercept, put, y = 0 in equation,
$\displaystyle 0 = \frac{2}{(x+1)(x-3)}$
we do not get any answer from this, so there is no, x-intercept. i.e., the graph does not cuts x-axis
Please see attached graph, the green lines show the asymptotes.