Question

Need the answers plz

Answer 1

1)
           x^2 + 4x - 5
y =   --------------------
              3x^2 - 12

Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite

=> 3x^2 - 12 = 0 => 3x^2 = 12

=> x^2 = 12 / 3 = 4

=> x = +/-2

Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) =  ∞

Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞

Limit of y when x -> - 2(+) = +∞

Limit of y when x -> - 2(-) = -

=> x = 2 and x = - 2 are a vertical asymptotes

Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite

Limi of y when x -> - ∞ = 1/3

Lim of y when x -> +∞ = 1/3

=> Horizontal asymptote y = 1/3

x - intercept => y = 0

=>
          x^2 + 4x - 5
0 =   ------------------ => x ^2 + 4x  - 5 = 0
              3x^2 - 12

Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1

=> x-intercepts x = - 5 and x = 1

Domain: all the real values except x = 2 and x = - 2

2)  y = - 2 / (x - 4) - 1

using the same criteria you get:

Vertical asymptote: x = 4

Horizontal asymptote: y = - 1

Domain:all the real values except x = 4

Range: all the real values except y = - 1

3) 

x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)

factor x^2 - 3x - 10 => (x - 5)(x + 2)

Multiply both sides by (x - 5) (x + 2)

=> x(x - 5) + 7( x + 2) = 14

=> x^2 - 5x + 7x + 14 = 14

=> x^2 + 2x = 0

=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0

Answer: x = 0