# [SOLVED] Finding polynomials from a diagram...

• August 2nd 2009, 01:31 PM
PTL
[SOLVED] Finding polynomials from a diagram...
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?
• August 2nd 2009, 01:50 PM
pickslides
as $x\neq -2,5$ then $q(x) = (x+2)(x-5)$

A horizontal asymptote could imply a constant term of +1 in $f(x)$

The points $f(-1) = f(4)=0$ will be used to solve any arbitrary constants.
• August 2nd 2009, 02:48 PM
PTL
Does that mean we have f(x) = $\frac{p(x)}{x^{2}-3x-10}$

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 $\frac{p(x)}{x^{2}-3x-10}$ 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)?
• August 2nd 2009, 02:56 PM
pickslides
There's a lot of Latex errors there my friend.
• August 3rd 2009, 01:18 PM
PTL
Quote:

Originally Posted by pickslides
There's a lot of Latex errors there my friend.

Fixed that, sorry...
• August 3rd 2009, 02:19 PM
DeMath
Quote:

Originally Posted by PTL
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?

If $\left[ \begin{gathered}f\left( { - 1} \right) = \frac{{p\left( { - 1} \right)}}{{q\left( { - 1} \right)}}=0, \hfill \\f\left( 4 \right) = \frac{{p\left( 4 \right)}}{{q\left( 4 \right)}} = 0; \hfill \\ \end{gathered} \right.$ then $\left[ \begin{gathered}p\left( { - 1} \right) = 0, \hfill \\p\left( 4 \right) = 0; \hfill \\ \end{gathered} \right. \Rightarrow p\left( x \right) = a\left( {x + 1} \right)\left( {x - 4} \right).$

So, we have

$f\left( x \right) = a \cdot \frac{{\left( {x + 1} \right)\left( {x - 4} \right)}}{{\left( {x + 2} \right)\left( {x - 5} \right)}} = a \cdot \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.$

If $y=1$ is a horizontal asymptote, then $\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1.$

Find the limit

$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = a \cdot\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}} = a \cdot\mathop {\lim }\limits_{x \to \infty }$ $\frac{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}}{{1 - {3 \mathord{\left/{\vphantom {3 x}} \right.\kern-\nulldelimiterspace} x} - {4 \mathord{\left/{\vphantom {4 {{x^2}}}} \right.\kern-\nulldelimiterspace} {{x^2}}}}} = a \cdot1.$

So $a = 1$

Finally $f\left( x \right) = \frac{{{x^2} - 3x - 4}}{{{x^2} - 3x - 10}}.$
• August 3rd 2009, 02:26 PM
Soroban
Hello, PTL!

Quote:

Find polynomials $p(x)$ and $q(x)$ such that: . $f(x) \:=\:\frac{p(x)}{q(x)}$ has vertical asymptotes $x=-2, x=5$,
horizontal asymptote $y=1$ and $f(-1) \:=\: f(4)\:=\:0$

For vertical asymptotes, the denominator must go to zero for $x = -2\text{ and }x = 5$
. . Hence: . $q(x) \:=\:(x+2)(x-5)$
So we have: . $f(x) \;=\;\frac{p(x)}{(x+2)(x-5)}$

For the horizontal asymptote, the function goes to 1 as $x \to \infty$
. . Hence, $p(x)$ has the same degree as $q(x)$, and has a leading coefficient of 1.
That is: . $p(x) \:=\:x^2 + ax + b$

And we have: . $f(x) \;=\;\frac{x^2+ax+b}{(x+2)(x-5)}$

Since $f(\text{-}1) \:=\:0\!:\quad f(\text{-}1) \;=\;\frac{(\text{-}1)^2 + a(\text{-}1) + b}{(\text{-}1+2)(\text{-}1-5)} \:=\:0 \quad\Rightarrow\quad a - b \:=\:1$ .[1]

Since $f(4) \:=\:0\!:\quad f(4) \;=\;\frac{4^2 + a(4) + b}{(4+2)(4-5)} \:=\:0 \quad\Rightarrow\quad 4a + b \:=\:-16$ .[2]

Add [1] and [2]: . $5a \:=\:-15 \quad\Rightarrow\quad \boxed{a \,=\,-3} \quad\Rightarrow\quad\boxed{ b \,=\,-4}$

Therefore: . $f(x) \;=\;\frac{x^2 - 3x - 4}{(x+2)(x-5)}$

Edit: Ah ... DeMath beat me to it.
Oh well, my explanation is slightly different.
.
• August 3rd 2009, 02:50 PM
DeMath