1. **State the problem:** We are given the function $f(x) = 3x^2 + x^3$ and asked to analyze its intercepts, symmetry, and graph behavior. 2. **Find x-intercepts where the graph touches and turns around:** Set $f(x) = 0$: $$3x^2 + x^3 = 0$$ Factor out $x^2$: $$x^2(3 + x) = 0$$ This gives $x^2 = 0$ or $3 + x = 0$. So, $x = 0$ or $x = -3$. At $x=0$, since the factor is $x^2$, the graph touches the x-axis and turns around (a repeated root). At $x=-3$, the factor is linear, so the graph crosses the x-axis. 3. **Check if there are other x-intercepts where the graph touches and turns around:** No, only at $x=0$ does the graph touch and turn around. 4. **Find the y-intercept:** Set $x=0$: $$f(0) = 3(0)^2 + (0)^3 = 0$$ So, the y-intercept is at $(0,0)$. 5. **Determine symmetry:** Check $f(-x)$: $$f(-x) = 3(-x)^2 + (-x)^3 = 3x^2 - x^3$$ Compare with $f(x)$: - $f(-x) \neq f(x)$, so no y-axis symmetry. - $f(-x) \neq -f(x)$, so no origin symmetry. Therefore, the graph has neither y-axis nor origin symmetry. **Final answers:** - A: The graph touches and turns around at $x=0$. - B: There are no other x-intercepts where the graph touches and turns around. - C: The y-intercept is $0$. - D: The graph has neither y-axis nor origin symmetry.