1. **State the problem:** We are given two quadratic functions: - $g(x) = 3x^2$ - $g(x) = \frac{1}{3}x^2$ We need to create tables of values, plot points, and determine the domain and range for each. 2. **Recall the formula and properties:** - Quadratic functions have the form $g(x) = ax^2$ where $a$ affects the steepness. - The domain of any quadratic function is all real numbers: $\text{Domain} = (-\infty, \infty)$. - The range depends on $a$: - If $a > 0$, the parabola opens upwards and the range is $[0, \infty)$. - If $a < 0$, it opens downwards. 3. **For $g(x) = 3x^2$:** - Make a table of values for $x = -2, -1, 0, 1, 2$: $$ \begin{aligned} g(-2) &= 3(-2)^2 = 3 \times 4 = 12 \\ g(-1) &= 3(-1)^2 = 3 \times 1 = 3 \\ g(0) &= 3(0)^2 = 0 \\ g(1) &= 3(1)^2 = 3 \\ g(2) &= 3(2)^2 = 12 \end{aligned} $$ - Domain: All real numbers $(-\infty, \infty)$. - Range: Since $a=3 > 0$, range is $[0, \infty)$. 4. **For $g(x) = \frac{1}{3}x^2$:** - Make a table of values for $x = -2, -1, 0, 1, 2$: $$ \begin{aligned} g(-2) &= \frac{1}{3}(-2)^2 = \frac{1}{3} \times 4 = \frac{4}{3} \\ g(-1) &= \frac{1}{3}(-1)^2 = \frac{1}{3} \\ g(0) &= 0 \\ g(1) &= \frac{1}{3} \\ g(2) &= \frac{4}{3} \end{aligned} $$ - Domain: All real numbers $(-\infty, \infty)$. - Range: Since $a=\frac{1}{3} > 0$, range is $[0, \infty)$. 5. **Summary:** Both functions have domain $(-\infty, \infty)$ and range $[0, \infty)$. 6. **Graph descriptions:** - $g(x) = 3x^2$ is a steeper parabola. - $g(x) = \frac{1}{3}x^2$ is wider and less steep. Final answers: - Domain for both: $(-\infty, \infty)$ - Range for both: $[0, \infty)$