1. **Problem Statement:** Identify which function among the options has a different range than the original function $f(x) = x^2$. 2. **Recall the range of $f(x) = x^2$:** The function $f(x) = x^2$ is a parabola opening upwards with vertex at $(0,0)$. Its range is all real numbers $y$ such that $$y \geq 0.$$ 3. **Analyze each option:** - a. $y = 4f(x) = 4x^2$ stretches the parabola vertically by a factor of 4. The vertex remains at $(0,0)$, so the range is $$y \geq 0.$$ - b. $y = f(x) + 4 = x^2 + 4$ shifts the parabola up by 4 units. The vertex moves to $(0,4)$, so the range is $$y \geq 4.$$ - c. $y = f(x + 4) = (x+4)^2$ shifts the parabola horizontally left by 4 units. The vertex moves to $(-4,0)$, but the minimum value remains 0, so the range is $$y \geq 0.$$ - d. $y = f(-x) = (-x)^2 = x^2$ reflects the parabola about the y-axis, but since $x^2$ is even, the graph is unchanged. The range remains $$y \geq 0.$$ 4. **Conclusion:** Only option b changes the range from $y \geq 0$ to $y \geq 4$. **Final answer:** Option b, $y = f(x) + 4$, has a different range.