1. Extra-credit problem

can anyone help me in graphing this ?

f(x) = whole sqrt x + 3/x - 2

2. Originally Posted by darkangel
can anyone help me in graphing this ?

f(x) = whole sqrt x + 3/x - 2
Is it f(x) = whole sqrt x + (3/x) - 2 or f(x) = whole sqrt x + 3/(x - 2)

3. i am sorry about that
its f(x) = whole sqrt (x + 3)/(x - 2)

4. Originally Posted by darkangel
its f(x) = whole sqrt (x + 3)/(x - 2)
$\displaystyle y = \sqrt{\frac{x+3}{x-2}} = \sqrt{\frac{[x-2]+5}{x-2}} = \sqrt{1 + \frac{5}{x-2}}$.

First of all you should establish the implied domain. You need $\displaystyle x + 3 \geq 0$ AND x - 2 > 0 => ......, OR $\displaystyle x + 3 \leq 0$ AND x - 2 < 0 => ......

There's no y-intercept (why?).

x-int: Solve x + 3 = 0 (why?).

Vertical asymptote is got from solving x - 2 = 0 (why?).

Horizontal asymptote: As $\displaystyle x \rightarrow \pm \infty$, y --> 1 (why?). In fact, as x --> -oo, y --> 1 from below (why?) and as x --> +oo, y --> 1 from above (why?)

Stationary points: Solve dy/dx = 0. (Hint: There are none).

Now draw the shape that ties all these features together .....

5. thank u so much for ur response...
umm i did not understand about the horizontal asymtote ( like how can i find it, if i get one of these problems on the test )

6. nevermind.. i figured it out