1. ## asymptote

ok so i've been given some isotopes and i'm not sure how to plot them, well i do but i'm kinda confuse .

ok so i've ben given this h(x) = 1/2x + 1
but the main one i'm confuse on its this -3 / 2x - 5 (totally dont get it)

like i know for the first one the vertical asotopes is -1/2 and the second one is 5/2. Cause i know that most of the time u have to draw the isotopes base on a given point. So how do u find that point? thanks!

2. Originally Posted by lickman
ok so i've been given some isotopes and i'm not sure how to plot them, well i do but i'm kinda confuse .

ok so i've ben given this h(x) = 1/2x + 1
but the main one i'm confuse on its this -3 / 2x - 5 (totally dont get it)

like i know for the first one the vertical asotopes is -1/2 and the second one is 5/2. Cause i know that most of the time u have to draw the isotopes base on a given point. So how do u find that point? thanks!
I believe you're talking about "asymptotes"

$h(x)=\frac{1}{2x+1}$

Look at the denominator. The denominator cannot be zero. So what would make the denominator zero?

$2x+1=0$
$x=-\frac{1}{2}$

So, we have a vertical asymptote at $x=-\frac{1}{2}$

In the other one, look at the denominator

$f(x)=\frac{-3}{2x-5}$

$2x-5=0$
$x=\frac{5}{2}$

Therefore, you have a vertical asymptote at $x=\frac{5}{2}$

Just graph the vertical lines through these points.

3. I think you mean h(x) = 1/(2x + 1)

An asymptote exists where the graph is undefined. The graph approaches but never reaches this line.

The graph is undefined when the denominator = 0 because we cannot divide by 0

so asymptote exists when 2x +1 = 0 : 2x = -1 : x = -1/2

in the second case an asymptote exists when 2x-5 = 0 : 2x = 5 : x=5/2

4. i get what u mean , cause we're learning about reciprocal of a linear function, i know how graph the vertical asymotopes, but the thing i dont get is the curve for a function like this y = 1/x, rite. cause i thought most of the time theres a given point, and then u just draw ur curve base it on it. So the problem is i dunno how to draw the curve, correctly. thanks

5. Originally Posted by lickman
i get what u mean , cause we're learning about reciprocal of a linear function, i know how graph the vertical asymotopes, but the thing i dont get is the curve for a function like this y = 1/x, rite. cause i thought most of the time theres a given point, and then u just draw ur curve base it on it. So the problem is i dunno how to draw the curve, correctly. thanks
Assign arbitrary values for x (except 0, of course) and use the function to solve for y. Then plot the points.

If x=1, then y=1
If x=2, then y= 1/2
If x=-1, then y=-1
.
.
.
etc.

6. When graphing you may choose values of x eg. x=1 or x = 2 and find their corresponding y value

if h(x) = 1/(2x+1)

then h(1) = 1/(2(1)+1) = 1/3

Note that for large x that the denominator becomes so large that h(x) approaches zero

For very small x then x approaches zero and h(x) above approaches 1