How do you graph #f(x)=-(x+4)/2# using holes, vertical and horizontal asymptotes, x and y intercepts?

Question

Julia Adams · Accepted Answer

This function is not a rational function but rather can be broken down.

(-x+4)/2 can be rewritten as - #1/2x+2#. This is a simple linear function without asymptotes or holes. An asymptote occurs when you have a (x) on the bottom of your function. The value that causes (x) to become zero and make the function undefined is your asymptote. Holes occur when terms containing (x) can be cancelled of from the numerator and the denominator. This causes that gap in your function when graphed. However neither of these 2 are in your example.

Avery Alexander · Answer

undefined To graph the function f(x) = -(x+4)/2, we can start by identifying the vertical and horizontal asymptotes, x and y intercepts, and any holes in the graph.

Vertical asymptote:
There is no vertical asymptote in this case.

Horizontal asymptote:
To find the horizontal asymptote, we can analyze the behavior of the function as x approaches positive or negative infinity. In this case, as x approaches positive or negative infinity, the function approaches -(∞+4)/2 = -∞/2 = -∞. Therefore, the horizontal asymptote is y = -∞.

X-intercept:
To find the x-intercept, we set y = 0 and solve for x. In this case, -(x+4)/2 = 0. By multiplying both sides by -2, we get x + 4 = 0. Solving for x, we find x = -4. Therefore, the x-intercept is (-4, 0).

Y-intercept:
To find the y-intercept, we set x = 0 and solve for y. In this case, -(0+4)/2 = -4/2 = -2. Therefore, the y-intercept is (0, -2).

Holes:
There are no holes in the graph of this function.

To summarize:
- There is no vertical asymptote.
- The horizontal asymptote is y = -∞.
- The x-intercept is (-4, 0).
- The y-intercept is (0, -2).
- There are no holes in the graph.