Question

Hey, Its the same person. This is Algebra 2. Please Help!

Answer 1

Answer:

first option

Step-by-step explanation:

Given

f(x) = $\frac{2x^2+5x-12}{x+4}$ ← factorise the numerator

     = $\frac{(x + 4)(2x-3)}{x+4}$ ← 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 = $\frac{3}{2}$

There is a zero at ( $\frac{3}{2}$, 0 )

Thus

discontinuity at (- 4, - 11 ), zero at ( $\frac{3}{2}$, 0 )