# Finding the equation of a rational function.

• Nov 8th 2009, 05:14 PM
thisisnew
Finding the equation of a rational function.
I need help with this problem as soon as possible.

Suppose you have the function f(x)=2x+b
cx+d
This function is undefined at x=-4
3, the graph of this function strikes the x axis at 8 and the y axis at 1. What are the values for b,c and d?

I have no idea how to do any of this
• Nov 8th 2009, 05:32 PM
I-Think
I'm assuming you mean $f(x)=\frac{2x+b}{cx+d}$

Hints
If the function is undefined at $x=\frac{-4}{3}$, it usually means a division by zero takes place. *hint*hint :)

If it strikes the x axis at 8, that means when f(x)=0, x=8, i.e.:f(8)=0
And if it strikes the y-axis at 1, that means that at x=0, f(x)=1, i.e. f(0)=1

Substitute the values and use the hints.
• Nov 8th 2009, 08:33 PM
thisisnew
I'm still not fully understanding this...ugh
• Nov 8th 2009, 08:54 PM
Bacterius
Alright, you have :

$f(x)=\frac{2x+b}{cx+d}$

You also know that a division per zero necessarily occurs in this function if $x \in R$, if and only if $c <> 0$. You can reasonably assume that $c$ is not equal to zero since that would be really boring. And when a division per zero occurs, what happens ? The function is undefined.

Therefore, if the function is undefined at $x = \frac{ -4}{3}$, the denominator is equal to $0$, which means you have to solve $\frac{ -4}{3} c + d = 0$

The graph of this function cuts the x-axis in $x = 8$, which means that when $f(x) = 0$, $x = 8$. You then have $f(8) = 0$.

The graph of this function cuts the y-axis in $y = 1$, which means that when $f(x) = 1$, $x = 0$. You then have $f(0) = 1$.

Let's resume all this :

$f(x)=\frac{2x+b}{cx+d}$

$\frac{ -4}{3}c + d = 0$

$f(8) = 0$ , equivalent to $\frac{2 \times 8 +b}{c \times 8+d} = 0$

$f(0) = 1$, equivalent to $\frac{b}{d} = 1$

______________________

Now I'll do a bit for you in this spoiler (don't click if you want to find by yourself) :

Spoiler:
$\frac{b}{d} = 1$ means that $b = d$

Therefore, the function becomes :

$f(x)=\frac{2x+b}{cx+b}$

Now you know that $\frac{ -4}{3} c + d = 0$ , which means that $c = \frac{ 3d}{4}$

Therefore, the function becomes :

$f(x)=\frac{2x+b}{\frac{ 3d}{4} x+b}$

Remember that $b = d$, so we have :

$f(x)=\frac{2x+b}{\frac{ 3b}{4} x+b}$

Now use one of the x-axis / y-axis values to substitute, isolate and solve $b$.

You can solve for $b$, therefore you can solve for $d$, and the last variable remaining ( $c$) can be found by reverting to the original function and substituting the y-axis cut values.

I might have done some errors in the spoiler though, so don't just copy blindly.
• Nov 8th 2009, 09:09 PM
ukorov
Quote:

Originally Posted by thisisnew
Suppose you have the function f(x)=2x+b
cx+d
This function is undefined at x=-4
3, the graph of this function strikes the x axis at 8 and the y axis at 1. What are the values for b,c and d?

I have no idea how to do any of this

when $x = -\frac{4}{3}$, cx + d = 0, hence
$d - \frac{4c}{3} = 0$
f(x) strikes x-axis at (8,0) and y-axis at (0,1)
hence $0 = \frac{2(8) + b}{c(8) + d}$ ,
hence 16 + b = 0
b = -16
also $1 = \frac{2(0) + b}{c(0) + d}$ ,
hence c(0) + d = 2(0) + b
0 + d = 0 + (-16)
d = -16
Since $d - \frac{4c}{3} = 0$ ,
$d = \frac{4c}{3}$
$c = (-16)(\frac{3}{4})$
c = -12
• Nov 8th 2009, 09:17 PM
Bacterius
Ukorov, you can simplify a bit your work by clearly saying that $b = d$, therefore getting rid of the part involving evaluation of $d$. It makes it a bit shorter, and a lot cooler I find (Happy)