b) \( f(x)=x^{2}+6 x+9 \)

It looks like you're working with the quadratic function: \[ f(x) = x^{2} + 6x + 9 \] Let's analyze and simplify this function. ### 1. **Factoring the Quadratic** To factor the quadratic expression, we look for two numbers that multiply to **9** (the constant term) and add up to **6** (the coefficient of the middle term). - The numbers **3** and **3** satisfy this because: \[ 3 \times 3 = 9 \quad \text{and} \quad 3 + 3 = 6 \] So, the factored form of the quadratic is: \[ f(x) = (x + 3)(x + 3) = (x + 3)^2 \] ### 2. **Vertex Form** The factored form \((x + 3)^2\) indicates that the vertex of the parabola represented by \(f(x)\) is at **\((-3, 0)\)**. ### 3. **Roots of the Function** Setting \(f(x) = 0\) to find the roots: \[ (x + 3)^2 = 0 \implies x = -3 \] This means the function has a **double root** at \(x = -3\). ### 4. **Graphical Interpretation** The graph of \(f(x)\) is a parabola that opens ** upwards** (since the coefficient of \(x^2\) is positive) and touches the x-axis at the point \((-3, 0)\). ### 5. **Derivative (Optional)** If you're interested in calculus aspects, the derivative of \(f(x)\) with respect to \(x\) is: \[ f'(x) = 2x + 6 \] Setting the derivative to zero to find critical points: \[ 2x + 6 = 0 \implies x = -3 \] This confirms that the vertex at \(x = -3\) is indeed the minimum point of the function. --- **Summary:** - **Factored Form:** \(f(x) = (x + 3)^2\) - **Vertex:** \((-3, 0)\) - **Roots:** \(x = -3\) (double root) If you have a specific aspect you'd like to delve into further, feel free to ask!