Find polynomials p(x) and q(x) such that f(x) = p(x)/q(x) has vertical asymptotes x=-2, x=5, horizontal asymptote y=1 and f(-1) = f(4)=0.
Any hints on how to do this?
Does that mean we have f(x) = $\displaystyle \frac{p(x)}{x^{2}-3x-10}$
So p(-1) = 0
p(4) = 0
and p(x)/q(x) has a local max at (1.5, almost(1))?
So the derivative of $\displaystyle \frac{p(x)}{x^{2}-3x-10}$ at 1.5 = 0? (1.5 because it's half-way between -2 and 5..
Is that how I'm supposed to solve for p(x)?
If $\displaystyle \left[ \begin{gathered}f\left( { - 1} \right) = \frac{{p\left( { - 1} \right)}}{{q\left( { - 1} \right)}}=0, \hfill \\f\left( 4 \right) = \frac{{p\left( 4 \right)}}{{q\left( 4 \right)}} = 0; \hfill \\ \end{gathered} \right.$ then $\displaystyle \left[ \begin{gathered}p\left( { - 1} \right) = 0, \hfill \\p\left( 4 \right) = 0; \hfill \\ \end{gathered} \right. \Rightarrow p\left( x \right) = a\left( {x + 1} \right)\left( {x - 4} \right).$
So, we have
$\displaystyle f\left( x \right) = a \cdot \frac{{\left( {x + 1} \right)\left( {x - 4} \right)}}{{\left( {x + 2} \right)\left( {x - 5} \right)}} = a \cdot \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.$
If $\displaystyle y=1$ is a horizontal asymptote, then $\displaystyle \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1.$
Find the limit
$\displaystyle \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = a \cdot\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}} = a \cdot\mathop {\lim }\limits_{x \to \infty }$$\displaystyle \frac{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}}{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}} = a \cdot1.$
So $\displaystyle a = 1$
Finally $\displaystyle f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.$
Hello, PTL!
For vertical asymptotes, the denominator must go to zero for $\displaystyle x = -2\text{ and }x = 5$Find polynomials $\displaystyle p(x)$ and $\displaystyle q(x)$ such that: .$\displaystyle f(x) \:=\:\frac{p(x)}{q(x)}$ has vertical asymptotes $\displaystyle x=-2, x=5$,
horizontal asymptote $\displaystyle y=1$ and $\displaystyle f(-1) \:=\: f(4)\:=\:0$
. . Hence: .$\displaystyle q(x) \:=\:(x+2)(x-5)$
So we have: .$\displaystyle f(x) \;=\;\frac{p(x)}{(x+2)(x-5)} $
For the horizontal asymptote, the function goes to 1 as $\displaystyle x \to \infty$
. . Hence, $\displaystyle p(x)$ has the same degree as $\displaystyle q(x)$, and has a leading coefficient of 1.
That is: .$\displaystyle p(x) \:=\:x^2 + ax + b$
And we have: .$\displaystyle f(x) \;=\;\frac{x^2+ax+b}{(x+2)(x-5)}$
Since $\displaystyle f(\text{-}1) \:=\:0\!:\quad f(\text{-}1) \;=\;\frac{(\text{-}1)^2 + a(\text{-}1) + b}{(\text{-}1+2)(\text{-}1-5)} \:=\:0 \quad\Rightarrow\quad a - b \:=\:1$ .[1]
Since $\displaystyle f(4) \:=\:0\!:\quad f(4) \;=\;\frac{4^2 + a(4) + b}{(4+2)(4-5)} \:=\:0 \quad\Rightarrow\quad 4a + b \:=\:-16 $ .[2]
Add [1] and [2]: .$\displaystyle 5a \:=\:-15 \quad\Rightarrow\quad \boxed{a \,=\,-3} \quad\Rightarrow\quad\boxed{ b \,=\,-4} $
Therefore: .$\displaystyle f(x) \;=\;\frac{x^2 - 3x - 4}{(x+2)(x-5)} $
Edit: Ah ... DeMath beat me to it.
Oh well, my explanation is slightly different.
.