Question

1. state the internal over which a parabola is positive if it has zeros at -4 and 4, and has an initial value of 5.

Explanation:

Step1: Write the parabola equation

Since the zeros are at \(x = - 4\) and \(x = 4\), the parabola can be written in factored form as \(y=a(x + 4)(x - 4)\). The initial value (when \(x = 0\)) is \(5\). Substitute \(x = 0\) and \(y=5\) into the equation: \(5=a(0 + 4)(0 - 4)\), so \(5=a\times4\times(-4)\), then \(a=-\frac{5}{16}\). The equation is \(y =-\frac{5}{16}(x + 4)(x - 4)=-\frac{5}{16}(x^{2}-16)\).

Step2: Analyze the sign of the parabola

The parabola \(y =-\frac{5}{16}(x + 4)(x - 4)\) is a quadratic function. The coefficient of \(x^{2}\) is negative (\(a =-\frac{5}{16}<0\)), so the parabola opens downwards. The parabola is positive (above the \(x -\)axis) between its two roots.

Answer:

\((-4,4)\)