Determine X Values for Positive Outcomes in y = -x² + 5x + 6

The following function is graphed below:

$y=-x^2+5x+6$

For which values of x is

$f(x)>0$ true?

To solve the problem of finding when $f(x) = -x^2 + 5x + 6 > 0$, follow these steps:

Step 1: Find the roots of the quadratic equation $-x^2 + 5x + 6 = 0$. Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = -1$, $b = 5$, and $c = 6$.

Step 2: Calculate the discriminant:

$b^2 - 4ac = 5^2 - 4(-1)(6) = 25 + 24 = 49$

Step 3: Solve for $x$:

$x = \frac{-5 \pm \sqrt{49}}{-2}$

Step 4: Simplify the roots:

$x = \frac{-5 \pm 7}{-2}$

This gives the solutions:

$x = \frac{-5 + 7}{-2} = -1 \quad \text{and} \quad x = \frac{-5 - 7}{-2} = 6$

Step 5: Evaluate intervals defined by these roots: $(- \infty, -1)$, $(-1, 6)$, and $(6, \infty)$.

Step 6: Test a sample point in each interval to check where $f(x) > 0$:

• For interval $(- \infty, -1)$, choose $x = -2$. $f(-2) = -(-2)^2 + 5(-2) + 6 = -4 - 10 + 6 = -8$ (negative)
• For interval $(-1, 6)$, choose $x = 0$. $f(0) = -(0)^2 + 5(0) + 6 = 6$ (positive)
• For interval $(6, \infty)$, choose $x = 7$. $f(7) = -(7)^2 + 5(7) + 6 = -49 + 35 + 6 = -8$ (negative)

Step 7: Conclude that the function is positive in the interval: $-1 < x < 6$.

Therefore, the solution is $-1 < x < 6$.