QuestionFor the quadratic function $f(x)=x^{2}+4 x-5$, answer parts (a) through ( $f$. (Type your answer in interval notation.) (f) Determine where $f(x)>0$ and where $f(x)<0$. Select the correct choice below and fill in the answer box (es) within your choice. (Type your answer in interval notation. Use integers or fractions for any numbers in the expression.) A. $f(x)>0$ on $\square$, and $f(x)$ is never negative $\square$ B. $f(x)>0$ on $\square$ , and $f(x)<0$ on $\square$ C. $f(x)<0$ on $\square$ ], and $f(x)$ is never positive

Studdy Solution
Test a point in each interval to determine the sign of $f(x)$:
- For $(-\infty, -5)$, choose $x = -6$: $$f(-6) = (-6)^2 + 4(-6) - 5 = 36 - 24 - 5 = 7$$ So, $f(x) > 0$ on $(-\infty, -5)$.
- For $(-5, 1)$, choose $x = 0$: $$f(0) = 0^2 + 4(0) - 5 = -5$$ So, $f(x) < 0$ on $(-5, 1)$.
- For $(1, \infty)$, choose $x = 2$: $$f(2) = 2^2 + 4(2) - 5 = 4 + 8 - 5 = 7$$ So, $f(x) > 0$ on $(1, \infty)$.
The intervals where $f(x) > 0$ and $f(x) < 0$ are:
$$f(x) > 0 \text{ on } (-\infty, -5) \cup (1, \infty)$$ $$f(x) < 0 \text{ on } (-5, 1)$$
The correct choice is B:
$$f(x) > 0 \text{ on } (-\infty, -5) \cup (1, \infty), \text{ and } f(x) < 0 \text{ on } (-5, 1)$$