Can someone help me to graph this ? f(x) = x-1/ sqrt x^2 -x - 2 Thanks!!!
W woooo New member Joined Mar 11, 2008 Messages 7 Mar 17, 2008 #1 Can someone help me to graph this ? f(x) = x-1/ sqrt x^2 -x - 2 Thanks!!!
L Loren Senior Member Joined Aug 28, 2007 Messages 1,299 Mar 17, 2008 #2 Re: Extra-credit problem As written f(x) = x-1/ sqrt x^2 -x - 2 means \(\displaystyle f(x)=x-\frac{1}{\sqrt{x^2}}-x-2\) Maybe you want to put some parenthesis in there to clarify what you really mean.
Re: Extra-credit problem As written f(x) = x-1/ sqrt x^2 -x - 2 means \(\displaystyle f(x)=x-\frac{1}{\sqrt{x^2}}-x-2\) Maybe you want to put some parenthesis in there to clarify what you really mean.
W woooo New member Joined Mar 11, 2008 Messages 7 Mar 17, 2008 #3 Re: Extra-credit problem i am sorry about that its (x-1)/ sqrt(x^2-x-2)
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Mar 17, 2008 #4 Re: Extra-credit problem it may help to note that the expression inside the radical will factor.
W woooo New member Joined Mar 11, 2008 Messages 7 Mar 17, 2008 #5 Re: Extra-credit problem yea i did that, what should i do after that ?
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,582 Mar 18, 2008 #6 woooo said: yea i did that, what should i do after that ? Click to expand... Pick an x-value. Plug it in. Find the corresponding y-value. Draw the dot. Repeat the process until you have enough dots to sketch the graph. :wink: Eliz.
woooo said: yea i did that, what should i do after that ? Click to expand... Pick an x-value. Plug it in. Find the corresponding y-value. Draw the dot. Repeat the process until you have enough dots to sketch the graph. :wink: Eliz.
skeeter Elite Member Joined Dec 15, 2005 Messages 3,216 Mar 18, 2008 #7 you should also note that the graph will have two vertical asymptotes, two horizontal asymptotes, and a region of x-values where f(x) is undefined.
you should also note that the graph will have two vertical asymptotes, two horizontal asymptotes, and a region of x-values where f(x) is undefined.
W woooo New member Joined Mar 11, 2008 Messages 7 Mar 18, 2008 #8 thank u so much for everyone's help :mrgreen: . I was able to figure it out