Finding X: Solve y = x² + 5x + 4 for Negative Output

Look at the following function:

$y=x^2+5x+4$

Determine for which values of $x$ the following is true:

$f(x) < 0$

To determine where the function $f(x) = x^2 + 5x + 4$ is less than zero, we will first factor the quadratic expression.

Step 1: Factor the quadratic function.

• The quadratic $x^2 + 5x + 4$ can be factored into $(x + 4)(x + 1)$.

Step 2: Find the roots of the quadratic equation.

• Set each factor equal to zero: $x + 4 = 0$ and $x + 1 = 0$.
• The solutions are $x = -4$ and $x = -1$.

Step 3: Determine the sign of the quadratic in the intervals defined by these roots.

• Consider the intervals $(-\infty, -4)$, $(-4, -1)$, and $(-1, \infty)$.
• Pick test points from each interval: for $(-\infty, -4)$, choose $x = -5$; for $(-4, -1)$, choose $x = -2$; for $(-1, \infty)$, choose $x = 0$.
• Evaluate the sign of $(x+4)(x+1)$ at each test point:
• At $x = -5$, $(x+4)(x+1) = (-1)(-4) = 4$ (positive).
• At $x = -2$, $(x+4)(x+1) = (2)(-1) = -2$ (negative).
• At $x = 0$, $(x+4)(x+1) = (4)(1) = 4$ (positive).
• Therefore, the function is negative in the interval $(-4, -1)$.

Consequently, the solution is $-4 < x < -1$.

The correct choice from the options given is choice 4.