Solve the Quadratic: Finding Positive Intervals in y=1/2x² + 4.6x

To solve the problem, we need to determine where the function $y = \frac{1}{2}x^2 + 4\frac{3}{5}x$ is positive.

First, rewrite the function by converting $4\frac{3}{5}x$ to an improper fraction: $\frac{23}{5}x$. Thus, the function becomes:

$y = \frac{1}{2}x^2 + \frac{23}{5}x$.

Next, we solve the inequality $y > 0$. First, find where $y = 0$:

$\frac{1}{2}x^2 + \frac{23}{5}x = 0$.

Factor the equation:

$x \left(\frac{1}{2}x + \frac{23}{5}\right) = 0$.

This gives us the roots $x = 0$ and $\frac{1}{2}x + \frac{23}{5} = 0$.

Solve for the second root:

$\frac{1}{2}x = -\frac{23}{5}$

$x = -\frac{23}{5} \times 2$

$x = -\frac{46}{5} = -9\frac{1}{5}$.

The roots are $x = 0$ and $x = -9\frac{1}{5}$.

The function $y$ is a parabola opening upwards (as $\frac{1}{2} > 0$).

Using the roots, test intervals to find where $y > 0$:

• Test an $x$ value less than $-9\frac{1}{5}$ (e.g., $x = -10$): $y$ is negative.
• Test an $x$ value between $-9\frac{1}{5}$ and $0$ (e.g., $x = -5$): $y$ is positive.
• Test an $x$ value greater than $0$: $y$ is positive.

The inequality $f(x) > 0$ holds for $-9\frac{1}{5} < x < 0$.

The correct choice matches option 3: $-9\frac{1}{5} < x < 0$.