+0

0
252
3

I'm still not sure how to do these and I'm getting confused

1.

2.

3.

Feb 5, 2018

#1
+101252
+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

Feb 5, 2018
#2
+101252
+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

Feb 5, 2018
#3
+101252
+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

Feb 5, 2018