1. **State the problem:** Find the vertex and evaluate the quadratic function $y=3x^2 - 30x + 67$ at specific points. 2. **Formula for vertex:** The vertex $x$-coordinate of a parabola $y=ax^2+bx+c$ is given by $$x = -\frac{b}{2a}$$ 3. **Calculate vertex $x$-coordinate:** Here, $a=3$, $b=-30$, so $$x = -\frac{-30}{2 \times 3} = \frac{30}{6} = 5$$ 4. **Calculate vertex $y$-coordinate:** Substitute $x=5$ into the function: $$y = 3(5)^2 - 30(5) + 67 = 3 \times 25 - 150 + 67 = 75 - 150 + 67 = -8$$ 5. **Vertex:** The vertex is at the point $(5, -8)$. 6. **Evaluate at $x=4$:** $$y = 3(4)^2 - 30(4) + 67 = 3 \times 16 - 120 + 67 = 48 - 120 + 67 = -5$$ So the point is $(4, -5)$. 7. **Evaluate at $x=6$:** $$y = 3(6)^2 - 30(6) + 67 = 3 \times 36 - 180 + 67 = 108 - 180 + 67 = -5$$ So the point is $(6, -5)$. 8. **Graph shape:** Since $a=3 > 0$, the parabola opens upwards. Final answers: - Vertex: $(5, -8)$ - Points: $(4, -5)$ and $(6, -5)$