1. **State the problem:** Sketch the graph of the polynomial function $$y = 3x^4 + 4x^3 + 1$$ and analyze its critical points. 2. **Find the value at $$x=0$$:** $$y(0) = 3(0)^4 + 4(0)^3 + 1 = 1$$ So the graph passes through the point $$(0,1)$$. 3. **Find the first derivative to locate critical points:** $$y' = \frac{d}{dx}(3x^4 + 4x^3 + 1) = 12x^3 + 12x^2$$ 4. **Factor the derivative:** $$y' = 12x^2(x + 1) = 0$$ Critical points occur where $$y' = 0$$, so: $$12x^2 = 0 \implies x=0$$ $$x + 1 = 0 \implies x = -1$$ 5. **Second derivative to determine concavity:** $$y'' = \frac{d}{dx}(12x^3 + 12x^2) = 36x^2 + 24x$$ 6. **Evaluate $$y''$$ at critical points:** - At $$x=0$$: $$y''(0) = 36(0)^2 + 24(0) = 0$$ (inconclusive, test further or use first derivative test) - At $$x=-1$$: $$y''(-1) = 36(-1)^2 + 24(-1) = 36 - 24 = 12 > 0$$, so $$x=-1$$ is a local minimum. 7. **Summary:** - The graph passes through $$(0,1)$$. - Critical points at $$x=0$$ and $$x=-1$$. - Local minimum at $$x=-1$$. - The function is a quartic polynomial with positive leading coefficient, so it tends to $$+\infty$$ as $$x \to \pm \infty$$. This analysis helps sketch the curve showing the shape and critical points.