SOLUTION: hey there, I really donīt know in which category this goes, but my teacher gave me this graph in class which he told me he would exonerate me from the next test if I could solve i

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: hey there, I really donīt know in which category this goes, but my teacher gave me this graph in class which he told me he would exonerate me from the next test if I could solve i      Log On


   

Question 944138: hey there,
I really donīt know in which category this goes, but my teacher gave me this graph in class which he told me he would exonerate me from the next test if I could solve it. he gave me the final answer, y=x+(1/(x+1)) , but he told me to show how I could get it from the graph. right now Iīm completely lost and I really need to not take that test since i donīt feel prepared. If you could please input the equation in a graphing calculator so you can see the graph and how to solve it. he gave me a hint of using y-intercepts, x-intercepts, domain, range, vertical and horizontal asymptotes, and the x and y limits. please help me and thanks.
Answer by josgarithmetic(39614) About Me  (Show Source):
You can put this solution on YOUR website!
You can use a graphing calculator yourself, or use a friend's, or use one in your classroom, or.... use Google.com. For this google feature, just enter the equation exactly how you wrote it. You will then left-click on the GO button of the browser and your first webpage listing should be that graph.

Inspecting just your equation without using any graphing tools, notice x is undefined for -1, so this will seem to be a vertical asymptote at x=-1. Think what happens unbounded to the left. y will approach the value of x, meaning there be slant asymoptote of y=x as x goes unbounded to the left. What happens as x increases to the right without bound? Same thing. The slant asymptote y=x.

Unsure about roots or zeros?
Convert the partial expressions into a single rational expression for the right member.
x%28x%2B1%29%2F%28x%2B1%29%2B1%2F%28x%2B1%29


%28x%28x%2B1%29%2B1%29%2F%28x%2B1%29

%28x%5E2%2Bx%2B1%29%2F%28x%2B1%29

You can use the general solution method for a quadratic equation to find "roots" or "zeros" for the numerator, and you should find that there is NO REAL root. This means y=x%2B1%2F%28x%2B1%29 will not touch nor cross the x-axis.