So it wants me to find a formula. I got ((-9)(x-1))/((x+3)(x-3)). The last part of the graph is not coming out correctly though. How do I alter the formula so that the right side of the graph flips over the x-axis?
So it wants me to find a formula. I got ((-9)(x-1))/((x+3)(x-3)). The last part of the graph is not coming out correctly though. How do I alter the formula so that the right side of the graph flips over the x-axis?
Hello, DasRabbit!
There are vertical asymptotes at $\displaystyle x = 3$ and $\displaystyle x = \text{-}3.$
$\displaystyle f(x)$ has $\displaystyle (x-3)(x+3)$ in the denominator.
With the "usual" vertical asymptote, the graph goes to $\displaystyle +\infty$ on one side
. . and to $\displaystyle \text{-}\infty$ on the other side (as it does at $\displaystyle x = \text{-}3$).
At $\displaystyle x = 3$, the graph goes to $\displaystyle +\infty$ on both sides.
This indicates that the factor $\displaystyle (x-3)$ in the denominator is to an even power.
. . Hence, the denominator has: .$\displaystyle (x-3)^2(x+3)$
Since there is a horizontal asymptote $\displaystyle (y = 0)$, the numerator has a degree less than 3.
. . Assume (for simplicity) that the numerator is linear.
We have: .$\displaystyle f(x) \:=\:\frac{ax+b}{(x-3)^2(x+3)}$
We want: $\displaystyle f(0) = \text{-}1\!:\;\frac{b}{(-3)^2(3)} \:=\:\text{-}1 \quad\Rightarrow\quad b \,=\,\text{-}27$
We want: $\displaystyle f(1) = 0\!:\;\frac{a+b}{(\text{-}2)^2(4)} \:=\:0 \quad\Rightarrow\quad\frac{a-27}{16} \:=\:0 \quad\Rightarrow\quad a \:=\:27$
Therefore: .$\displaystyle f(x) \:=\:\frac{27x - 27}{(x-3)^2(x+3)} \quad\Rightarrow\quad f(x) \:=\:\frac{27(x-1)}{(x-3)^2(x+3)}$
Test a few points to ensure that graph is on the correct sides of the x-axis.