How do you graph #f(x)=(2x^2 + 5) / (x-1)#?

Question

Kennedy Adams · Accepted Answer

undefined To graph the function f(x) = (2x^2 + 5) / (x-1), follow these steps:

1. Determine the domain of the function by finding the values of x for which the denominator (x-1) is equal to zero. In this case, x cannot be equal to 1.

2. Find the vertical asymptotes by setting the denominator equal to zero and solving for x. In this case, x = 1 is a vertical asymptote.

3. Determine the behavior of the function as x approaches positive and negative infinity. Divide the leading term of the numerator (2x^2) by the leading term of the denominator (x) to find the horizontal asymptote. In this case, the horizontal asymptote is y = 2x.

4. Find the x-intercepts by setting the numerator equal to zero and solving for x. In this case, there are no x-intercepts.

5. Find the y-intercept by evaluating the function at x = 0. In this case, the y-intercept is (0, 5).

6. Plot the vertical asymptote, horizontal asymptote, x-intercepts (if any), and the y-intercept on the coordinate plane.

7. Choose additional x-values to evaluate the function and plot the corresponding points on the graph.

8. Connect the plotted points smoothly to form the graph of the function.

Note: It may be helpful to use a graphing calculator or software to visualize the graph accurately.

Eva Adamson · Answer

graph{2(x+1)+7/(x-1) [-20, 20, -20, 20]}

I can suggest to transform the function (defined everywhere but a point #x=1# when the denominator equals to zero) into the following form: #f(x) = (2x^2+5)/(x-1) = (2x^2-2+7)/(x-1) = [2(x-1)(x+1)+7]/(x-1)=# #= 2(x+1)+7/(x-1)# In this form we can graph separately #y = 2(x+1) = 2x+2# and #y=7/(x-1)#, after which we just add two graphs. #y=2x+2# graph{2x+2 [-10, 10, -5, 5]} #y=7/(x-1)# graph{7/(x-1) [-10, 10, -5, 5]} The sum of these graphs is #y= 2(x+1)+7/(x-1)# graph{2(x+1)+7/(x-1) [-20, 20, -20, 20]} Notice that #x=1# is an asymptote of this graph