#1**+1 **

(1) f(x) = 3 / [5x^5]

If x = 0, the denominator is undefined.....so....the function is defined at all values except at x = 0

So....the domain is (-inf, 0), U (0, inf)

We have a lower power polynomial in the numerator than in the denominator.....in a Lower/Higher situation, a rational function like this will always have a horizontal asymptote at y = 0

Also....because the power on the variable in the denominator is odd, the function will generate both negative and positve values in infinite directions both ways

So.....the range is (-inf, 0) U (0, inf)

Here's a graph to check this : https://www.desmos.com/calculator/4s1yykzw1v

CPhill
Feb 5, 2018

#2**+1 **

(2)

f(x) = 3 / [ x - 2 ] + 9

The function will be undefined when x = 2 [ it makes a denominator = 0]

Also...... we have a lower power polynomial over a higher power polynomial in the first fraction..... like the first problem, this sets up a horiontal asymptote at y = 0

However, the appended "9" shifts this asymptote* up* 9 units

So....we will have a horizontal asymptote at y = 9

Here's a graph to confirm our suspicions : https://www.desmos.com/calculator/p9km2zdfff

CPhill
Feb 5, 2018

#3**+1 **

(3)

f(x) = [ x^2 - 49] / [ x + 7] factor the denominator

f(x) = [ ( x + 7) ( x - 7) ] / ( x + 7) this simplifies to

f(x) = x - 7

This is a line with a y intercept at -7

Also...... x = -7 will make the original denominator 0, so this is will the "hole" will occur

So.....the first graph is correct

CPhill
Feb 5, 2018