Pre calc help!! please!?

Question

1) Given f(x) = (x - 4)/(x - 3) and g(x) = 2 - 7x, find (f - g)(x)

2) If f(x) = 3x - 1 and g(x) = 1/(2x - 1) find [f o g](x).

3) Given that x is an integer and -2 < x < 2, state the relation represented by y = |-4x| - 6 by listing a set of ordered pairs. Then state whether the relation is a function.

4) Write the equation of the line parallel to 3x - 15y + 35 = 0 and passing through (-4, -1).

5) Write the equation of the line perpendicular to y = (-1/3)x + 7/2 and passing through (2, 5).

Answer 1

1) The functions are

Answer 2

2) The functions are

             

            

           

          

         

Answer 3

4) The line equation is 3x - 15y + 35 = 0

Write the equation in slope - intercept form y = mx + b.

- 15y = - 3x - 35

y = (-3/-15)x - (35/-15)

y = (1/5)x + (7/3)

Comapare the equation with slope - intercept form.

Slope (m) = 1/5

Because the parallel lines have same slopes, the slope of parallel line through the point (- 4, - 1) is 1/5.

Now the parallel line equation is y = (1/5)x + b.

Find the y - intercept by substituting the point in the parallel line equation say (x, y) = (- 4, - 1).

- 1  = (1/5)(- 4) + b

b = - 1 + 4/5

b = (- 5 + 4)/5

⇒ b = - 1/5.

The parallel line equation is y = (1/5)x - (1/5).

Answer 4

5) The line equation y = (- 1/3)x + (7/2)

Comapare the equation with slope - intercept form y = mx + b.

Slope (m) = -1/3

Because the perpendicular lines  slopes are negitive reciprocal to each other, the slope of perpendicular line through the point (2, 5) is 3.

Now the perpendicular line equation is y = 3x + b.

Find the y - intercept by substituting the point in the perpendicular line equation say (x, y) = (2, 5).

5  = 3(2) + b

b =  - 6 + 5

⇒ b = - 1.

The perpendicular line equation is y = 3x - 1.

Answer 5

3)

- 2 < x < 2

The relation y = | 4x | - 6

• For x = -1

y = |4(- 1)| - 6

y = - 2

• For x = 0

y = | 4x | - 6

y = |4(0)| - 6

y = - 6

• For x = 1

y = | 4x | - 6

y = |4(1)| - 6

y = - 2

Now we can able to write the relation is { (-1,- 2), (0,- 6), (1, - 2)}

Domain set is {-1, 0, 1}

Range set is { -2, -6, -2}

The relation is  a function because the each element of the domain is paired with exactly one element of the range.