1. **State the problem:** We have a piecewise function defined as $$f(x) = \begin{cases} k(x^2 + 2x), & 1 \leq x \leq 2 \\ 0, & \text{elsewhere} \end{cases}$$ and we want to understand its behavior and graph. 2. **Formula and rules:** The function is zero outside the interval $[1,2]$ and inside this interval it is a quadratic expression multiplied by a constant $k$. The quadratic part is $x^2 + 2x$. 3. **Intermediate work:** Let's analyze the quadratic inside the interval: $$x^2 + 2x = x(x + 2)$$ At $x=1$, $$f(1) = k(1^2 + 2 \cdot 1) = k(1 + 2) = 3k$$ At $x=2$, $$f(2) = k(2^2 + 2 \cdot 2) = k(4 + 4) = 8k$$ 4. **Explanation:** The function is zero everywhere except between $x=1$ and $x=2$, where it follows a parabola scaled by $k$. The parabola opens upwards since the coefficient of $x^2$ is positive. 5. **Graph shape:** The graph is zero outside $[1,2]$ and rises from $3k$ at $x=1$ to $8k$ at $x=2$ following the quadratic curve. 6. **Additional functions given:** - $y = x^3 \cos 2x$ is a cubic modulated by cosine. - $y = (2x^2 - 2)^4$ is a quartic of a quadratic expression. - $y = 5x^3 - 7x^2 + 4x + 4$ is a cubic polynomial. These are separate functions and not part of the piecewise $f(x)$. **Final answer:** The piecewise function $f(x)$ is zero outside $[1,2]$ and equals $k(x^2 + 2x)$ inside $[1,2]$, with values $3k$ at $x=1$ and $8k$ at $x=2$.