PLEASE I NEED HELPE WITH THESE 5 Problem

Question

PROBLEM QUESTION
1.f(x)=(2x-8)/(3x-12)(2x-8) 1.Determine discontinuity and domain

2.f(x)=x-3/3x^2-7x-6 2.The discontinuities exist at x= and x=
The hole in the graph exist at x=
The vertical asymptote of the graph is at x

3.f(x)=x-3/3x^2-7x-6 3.The degree of the numerator is equal to
The degree of denominator is equal to
The equation for the horizontal asymptote of the function is

4.f(x)=(2x-5)(3x-2)/(2x-5)(3x-9) 4. The hole in the graph is at x=
While the vertical asymptote is at x=

5.f(x)=x-3/3x^2-7x-6 5.The y intercept is at (0,?)
The x-intercept is at (?,0)

Answer 1

A discontinuity is the value of x where the function does not exist.  All the x values where the function does exist is the domain.

 

1)

 

f(x) = (2x - 8) / [(3x - 12)(2x - 8)]

 

 

We set the denominator equal to zero.  This is because when we divide by zero, the function is undefined.

 

(3x - 12)(2x - 8) = 0

 

 

Set the factor equal to zero.

 

3x - 12 = 0             and           2x - 8 = 0

 

3x = 12                   and           2x = 8

 

 x = 4                    and              x = 4

 

 

The discontinuity happens at  x=4.

 

The domain is  (-∞, 4)∪(4, ∞)

 

 

2)

 

f(x) = (x - 3) / [(3x² - 7x - 6)²]

f(x) = (x - 3) / [((3x + 2)(x - 3))²]

 

 

When we find the discontinuity, we also find the vertical asymptote.

 

Set the denominator to zero.

 

3x + 2 = 0              and             x - 3 = 0

 

3x = -2                    and           x = 3

 

  x = -2/3                

 

 

The vertical discontinuity and vertical asymptote is  x=-2/3   and    x=3.

 

To find the hole, we expand f(x), since the denominator is squared.

 

f(x) = 1 / [(3x + 2)(3x +2)(x - 3)]

 

From this newly written function, there is no hole.

 

 

4)  is the same as problem 2.

 

 

 

5)

 

f(x) = (x - 3) / [(3x² - 7x - 6)⁵]

 

 

To find the y-intercept, set x=0.

 

f(x) = -3 / (-6)⁵

f(x) = -6 / -7776

f(x) = 6/7776

 

The y-intercept is (0, 6/7776).

 

 

To find the x-intercept, set the numerator equal to zero.

 

x - 3 = 0

x = 3

 

But wait. The denominator in this function is the same as the one in problem 2. The only difference is that they are expanded differently.  We found in that problem that the vertical asymptote is x=3.  Therefore, this is no x-intercept because the function will never touch x=3.