Answer:
first option
Step-by-step explanation:
Given
f(x) = [tex]\frac{2x^2+5x-12}{x+4}[/tex] ← factorise the numerator
= [tex]\frac{(x + 4)(2x-3)}{x+4}[/tex] ← cancel (x + 4) on numerator/ denominator
= 2x - 3
Cancelling (x + 4) creates a discontinuity ( a hole ) at x + 4 = 0, that is
x = - 4
Substitute x = - 4 into the simplified f(x) for y- coordinate
f(- 4) = 2(- 4) - 3 = - 8 - 3 = - 11
The discontinuity occurs at (- 4, - 11 )
To obtain the zero let f(x) = 0, that is
2x - 3 = 0 ⇒ 2x = 3 ⇒ x = [tex]\frac{3}{2}[/tex]
There is a zero at ( [tex]\frac{3}{2}[/tex], 0 )
Thus
discontinuity at (- 4, - 11 ), zero at ( [tex]\frac{3}{2}[/tex], 0 )
