Precalculus - another question!?

Question

Find the x- and y-intercepts of f(x) = (4 - x) / x, then determine whether the graph of f touches or crosses the x-axis at the x-intercept.

Can someone please show answer, and explain in detail how to get this answer?

Answer 1

We find the x-intercept by setting y equal to zero, and we find the y-intercept by setting x equal to zero. Where y = 0, we determine if x touches or crosses the x-axis by finding the first and second derivative of f(x). If the first derivative is zero and the second derivative is non-zero, it only touches. If the first derivative is non-zero, or the first and second derivative are both zero, it crosses.

To find the y-intercept, set x = 0. This is, evaluate f(0). We get (4 - 0) / 0, which is undefined. The function is undefined at x = 0, so there is no y-intercept. To find the x-intercept, set y = 0. 0 = (4 - x) / x ==> 4 - x = 0 ==> x = 4. It's good to check the value when you have tricky things like x in the denominator. f(4) = (4 - 4) / 4 = 0/4 = 0. So the x-intercept is 4.

f'(x) = (x(-1) - (4 - x)(1)) / x^2 = (-x - 4 + x) / x^2 = -4 / x^2
f'(4) = -4 / 4^2 = -4/16 = -0.25

This is non-zero, so the function crosses the x-axis at the x-intercept. We don't need to find f''(x).

Answer 2

It always helps to simplify the function before you begin. (4-x)/x is simplified to 4/x - 1. The graph of f(x)=k/x, where k is any constant real number, looks like a hyperbola on the diagonal and in the first and third quadrants. Asymptotes are y=0 and x=0. Now, we have 4/x - 1 so this means a vertical downshift of exactly one unit for the entire graph. Therefore what would normally be an asymptote at x=0 is now at x= -1. Visually, this indeed does cross the x-axis now as the function slopes negatively out of the first quadrant and into the fourth.
If you want to find the exact value of the x-intercept to prove that the drawing/quick theory is correct, solve for f(x) = 0:
0 = 4/x -1.
1 = 4/x
x = 4

Answer 3

Ther are no y-intercepts because f(x) is discontinuous at x = 0. And x = 0 is where y-intercept occurs.

When f(x) = 0 we have (4-x)/x = 0 --> x=4.
So x=4 is the x -intercept

Set x = 5 and yo get f(x ) = (4-5)/5 = -1/5 = -0.2
Therefore f(x) definitely crosses the x-axis.

Answer 4

You get the y intercept when x=0 and the x intercept when f(x)=y=0.

f(x)=0 --> (4-x)/x = 0 --> x=4 at y=0

y=f(0)=4/0 which is undefined thus the graph does not touch the x axis as x=0 is not part of the domain.