Thread: Determin an equation from a rational fucntion graph

Please Help!!

Basically the graph is showing:

vertical asymptote (V)on x=3
and horizontal asymptote(H) on x=2
and a point in the graph (1.5,0)

So this is how I got an answer
y=ax+b
------
cx+d
since H=a/c a=2c
V=-d/c d=-3c
to find b, i did some substitution

y=2cx+b
-------
cx-3c

substitute a point from the graph

0=2c(1.5)+b
-----------
c(1.5)-3c
0=3c+b
-----------
-1.5c
therefore
b=-4.5 c

I sub in b into the general equation

y=2cx-4.5c
----------
cx-3c
y= c(2x-4.5)
---------- c cancel out
c(x-3)

final answer is
y=2x-4.5
------
x-3

but my book's answer is

y=2x-3
---------
x-3


if it helps im using nelson advanced functions book 2008
pg 154 question 7

please help!! and if anyone can point me to a website that has better explaining in terms of rational functions in the form


y=ax+b
------
cx+d

Originally Posted by jepal

Basically the graph is showing:

vertical asymptote (V)on x=3
and horizontal asymptote(H) on x=2
and a point in the graph (1.5,0)

If the vertical asymptote is at x = 3, then x - 3 is a factor of the denominator, and no other such linear (neatly solveable in the whole numbers) factor exists there.

If the function has an horizontal asymptote that is off the x-axis, then the degree of the numerator is the same as the degree of the denominator.

The simplest conclusion would then be that the rational function is of the form f(x) = (2x + b)/(x - 3), where "b" is some constant.

We are given that f(1.5) = 0, so (3 + b)/(1.5 - 3) = 0. A fraction is zero only when its numerator is zero, so 3 + b = 0.

What then is the value of "b"?

Originally Posted by stapel

If the vertical asymptote is at x = 3, then x - 3 is a factor of the denominator, and no other such linear (neatly solveable in the whole numbers) factor exists there.

If the function has an horizontal asymptote that is off the x-axis, then the degree of the numerator is the same as the degree of the denominator.

The simplest conclusion would then be that the rational function is of the form f(x) = (2x + b)/(x - 3), where "b" is some constant.

We are given that f(1.5) = 0, so (3 + b)/(1.5 - 3) = 0. A fraction is zero only when its numerator is zero, so 3 + b = 0.

What then is the value of "b"?

ok thanks for the reply but im still a bit confuse. In terms of f(x)=(2x+b), where did you get the 2 from or how did u go about determing the two. Im assuming its the horizontal asymptote. Will this be case all the time, if the horizontal asymptote is off the x-axis.

Originally Posted by jepal

In terms of f(x)=(2x+b), where did you get the 2 from or how did u go about determing the two.

What is the horizontal asymptote?

How is this value obtained?

Given a (simplest) denominator of "x - 3", what must the leading coefficient of the numerator be?

Originally Posted by jepal

Will this be case all the time, if the horizontal asymptote is off the x-axis.

No. Review the rules for asymptotes; in particular, what happens when the degree of the denominator is greater than the degree of the numerator.