Question

to sketch the graph of a rational function, follow the following steps:
step 1 write the polynomial in the numerator and denominator in factored form
step 2 identify holes
step 3 identify vertical asymptotes
step 4 identify horizontal or slant asymptotes
step 5 establish intervals on the x - axis where either an x - intercept or a discontinuity occurs and then check the signs of the factors on those intervals to determine whether the graph lies above or below the x - axis.
testing intervals
to test if the graph lies above or below the x - axis, create a number line with intervals. then choose test values before, in between, and after the intervals to determine if the function is positive or negative in that area. a positive value means the graph is above the axis, and a negative value means the graph is below the axis.
try this - practice testing values to determine if the values are above or below the axis. circle if the graph of f(x) would be positive (+) or negative (-) in areas a, b, c, and d. area d is done for you below.
f(x)=\frac{x + 2}{(x + 1)(x - 3)}
match the appropriate test values for each area.
x = 10quad x=-1.5quad x=-5quad x = 0
area a
area b
area c
area d x = 10
ightarrow f(10)=\frac{12}{(11)(7)}=\frac{12}{77} positive (+)

Explanation:

Step1: Determine intervals

The function $f(x)=\frac{x + 2}{(x + 1)(x - 3)}$ has vertical asymptotes at $x=-1$ and $x = 3$ and an x - intercept at $x=-2$. The intervals are $(-\infty,-2)$, $(-2,-1)$, $(-1,3)$ and $(3,\infty)$.

Step2: Match test - values to intervals

Area A is $(-\infty,-2)$. A test - value in this interval is $x=-5$.
Area B is $(-2,-1)$. A test - value in this interval is $x=-1.5$.
Area C is $(-1,3)$. A test - value in this interval is $x = 0$.
Area D is $(3,\infty)$ and the given test - value is $x = 10$.

Step3: Find signs of function at test - values

For $x=-5$: $f(-5)=\frac{-5 + 2}{(-5 + 1)(-5 - 3)}=\frac{-3}{(-4)(-8)}=-\frac{3}{32}$, so negative (-).
For $x=-1.5$: $f(-1.5)=\frac{-1.5 + 2}{(-1.5 + 1)(-1.5 - 3)}=\frac{0.5}{(-0.5)(-4.5)}=\frac{0.5}{2.25}$, so positive (+).
For $x = 0$: $f(0)=\frac{0+2}{(0 + 1)(0 - 3)}=\frac{2}{(1)(-3)}=-\frac{2}{3}$, so negative (-).
For $x = 10$: $f(10)=\frac{10 + 2}{(10 + 1)(10 - 3)}=\frac{12}{(11)(7)}=\frac{12}{77}$, so positive (+).

Answer:

	Match the appropriate test values for each area	Sign of $f(x)=\frac{x + 2}{(x + 1)(x - 3)}$
Area B	$x=-1.5$	positive (+)
Area C	$x = 0$	negative (-)
Area D	$x = 10$	positive (+)