How do you graph #f(x)=3/(x+1)-2# using holes, vertical and horizontal asymptotes, x and y intercepts?

Question

Eva Adamson · Accepted Answer

graph{3/(x+1)-2 [-10, 10, -5, 5]}

As you can see, there is a vertical asymptote at about #y=1#(because you can never get 1 out of this equation), and a horizontal asymptote at about #x=-1# (because plugging in #-1# for #x# creates a fraction with zero on the bottom, specifically #3/0#). There is an #x#-intercept at #y=1/2#, and a #y#-intercept at #x=1#. There are no holes.

Genesis Allen · Answer

undefined To graph the function f(x) = 3/(x+1) - 2, we can start by identifying the vertical and horizontal asymptotes, holes, x-intercepts, and y-intercept.

Vertical asymptote: The function has a vertical asymptote at x = -1, since the denominator (x+1) becomes zero at that point.

Horizontal asymptote: As x approaches positive or negative infinity, the function approaches y = -2. Therefore, y = -2 is the horizontal asymptote.

Hole: There is a hole in the graph at x = -1, since the function is undefined at that point.

X-intercept: To find the x-intercept, we set y = 0 and solve for x. In this case, 0 = 3/(x+1) - 2. Solving this equation, we get x = -4/3 as the x-intercept.

Y-intercept: To find the y-intercept, we set x = 0 and solve for y. Plugging x = 0 into the function, we get y = 3/(0+1) - 2 = 1 - 2 = -1. Therefore, the y-intercept is -1.

To graph the function, plot the vertical asymptote at x = -1, the hole at (-1, undefined), the horizontal asymptote at y = -2, the x-intercept at (-4/3, 0), and the y-intercept at (0, -1).