1. **State the problem:** Graph the quadratic function $g(x) = -3x^2$, plot points, sketch the curve, and find domain and range. 2. **Make a table of values:** Choose $x$ values: $-2, -1, 0, 1, 2$. Calculate $g(x)$: $$g(-2) = -3(-2)^2 = -3 \times 4 = -12$$ $$g(-1) = -3(-1)^2 = -3 \times 1 = -3$$ $$g(0) = -3(0)^2 = 0$$ $$g(1) = -3(1)^2 = -3$$ $$g(2) = -3(2)^2 = -12$$ 3. **Plot points and sketch curve:** Points are $(-2, -12), (-1, -3), (0, 0), (1, -3), (2, -12)$. The parabola opens downward because the coefficient of $x^2$ is negative. 4. **Domain and range:** - Domain of any quadratic is all real numbers: $\boxed{(-\infty, \infty)}$ - Range is all $y$ values less than or equal to the vertex value (maximum here at $y=0$): $\boxed{(-\infty, 0]}$ --- 1. **State the problem:** Graph the quadratic function $g(x) = -\frac{1}{3}x^2$, plot points, sketch the curve, and find domain and range. 2. **Make a table of values:** Choose $x$ values: $-3, -1.5, 0, 1.5, 3$. Calculate $g(x)$: $$g(-3) = -\frac{1}{3}(-3)^2 = -\frac{1}{3} \times 9 = -3$$ $$g(-1.5) = -\frac{1}{3}(-1.5)^2 = -\frac{1}{3} \times 2.25 = -0.75$$ $$g(0) = 0$$ $$g(1.5) = -\frac{1}{3}(1.5)^2 = -0.75$$ $$g(3) = -3$$ 3. **Plot points and sketch curve:** Points are $(-3, -3), (-1.5, -0.75), (0, 0), (1.5, -0.75), (3, -3)$. The parabola opens downward and is wider than the first because the coefficient magnitude is smaller. 4. **Domain and range:** - Domain is all real numbers: $\boxed{(-\infty, \infty)}$ - Range is all $y$ values less than or equal to the vertex value $0$: $\boxed{(-\infty, 0]}$