Question

(50 points!) Consider the function f(x) = 3x-1/x+4.

Answer 1

we are given

$f(x)=\frac{3x-1}{x+4}$

(a)

We know that when denominator of any function is 0

then function will become undefined

so, denominator=0 will make function  no solution

so, we set denominator =0

and then we solve for x

$x+4=0$

$x=-4$............Answer

(b)

Vertical shift:

Suppose, we want to shift y=f(x) function vertically by 'c' units

we add 'c' top y-value

so, new function will become

$y=f(x)+c$

now, we have

If g(x) is a vertical shift of 4 units of f(x)

so, we can write our function as

$g(x)=f(x)+4$

we can plug f(x)

$g(x)=\frac{3x-1}{x+4}+4$

Comparison:

So, graph of f(x) is moved upside by 4 units to get graph of g(x)

(c)

We can set g(x)=8

and then we can solve for x

$g(x)=\frac{3x-1}{x+4}+4=8$

Multiply both sides by x+4

$\frac{3x-1}{x+4}\left(x+4\right)+4\left(x+4\right)=8\left(x+4\right)$

$3x-1+4\left(x+4\right)=8\left(x+4\right)$

$7x+15=8x+32$

$7x=8x+17$

$-x=17$

$x=-17$..............Answer