Question

what are the discontinuity and zero of the function f(x) = (x² + 5x + 4)/(x + 4)? (5 points) discontinuity at (4,3), zero at (1,0) discontinuity at (4,3), zero at (-1,0) discontinuity at (-4,-3), zero at (1,0) discontinuity at (-4,-3), zero at (-1,0)

Explanation:

Step1: Factor the numerator

Factor $x^{2}+5x + 4$ as $(x + 1)(x+4)$. So $f(x)=\frac{(x + 1)(x + 4)}{x + 4}$.

Step2: Find the zero - point

Set the numerator equal to zero after canceling out common factors (excluding the factor that causes the discontinuity). After canceling $x + 4$ (for $x
eq - 4$), we set $x+1=0$, and we get $x=-1$ as the zero - point of the simplified function.

Step3: Find the discontinuity

The function $f(x)=\frac{x^{2}+5x + 4}{x + 4}$ is undefined when the denominator is zero. Set $x + 4=0$, we get $x=-4$. When $x=-4$, $y=\frac{(-4)^{2}+5\times(-4)+4}{-4 + 4}=\frac{16-20 + 4}{0}$, which is undefined. Substituting $x=-4$ into the original numerator gives $y$ - value of the non - simplified function. The function has a discontinuity at $x=-4$. To find the $y$ - value of the non - simplified function when $x=-4$, we can consider the behavior. The function has a discontinuity at $x=-4$. When $x=-4$, we can analyze the original function. The function $f(x)$ has a discontinuity at $x=-4$. If we consider the limit as $x\to - 4$, we can rewrite the function as $f(x)=x + 1$ for $x
eq-4$. The value of the function approaches $-4 + 1=-3$ as $x\to - 4$. So the discontinuity is at $(-4,-3)$.

Answer:

Discontinuity at $(-4,-3)$, zero at $(-1,0)$