Answer:


Explanation:
Question 1
The intervals on which a quadratic function is positive are those intervals where the function is above the x-axis, i.e. where y > 0.
The zeros of the quadratic function are the points at which the parabola crosses the x-axis.
As the given quadratic function has a negative leading coefficient, the parabola opens downwards. Therefore, the interval on which y > 0 is between the zeros.
To find the zeros of the given quadratic function, substitute y = 0 and factor:

Apply the zero-product property:


Therefore, the interval on which the function is positive is:
- Solution: 1 < x < 4
- Interval notation: (1, 4)
Question 2
The intervals on which a quadratic function is negative are those intervals where the function is below the x-axis, i.e. where y < 0.
The zeros of the quadratic function are the points at which the parabola crosses the x-axis.
As the given quadratic function has a positive leading coefficient, the parabola opens upwards. Therefore, the interval on which y < 0 is between the zeros.
To find the zeros of the given quadratic function, substitute y = 0 and factor:

Apply the zero-product property:


Therefore, the interval on which the function is negative is:
- Solution: -2 < x < 4
- Interval notation: (-2, 4)