Question: Determine the vertex and direction of opening of the parabola for the following quadratic equation: $$y = 3x^2 - 18x - 10$$
a.) Vertex: (3,-37) Opens: Upward
b.) Vertex: (-3,71) Opens: Downward
c.) Vertex: (-3,71) Opens: Upward
d.) Vertex: (3,-37) Opens: Downward
1. **State the problem:** Find the vertex and direction of opening of the parabola given by $$y = 3x^2 - 18x - 10$$.
2. **Recall the vertex formula:** For a quadratic equation in the form $$y = ax^2 + bx + c$$, the vertex $$ (h, k) $$ is found by:
$$ h = -\frac{b}{2a} $$
$$ k = f(h) = a h^2 + b h + c $$
3. **Identify coefficients:** Here, $$a = 3$$, $$b = -18$$, and $$c = -10$$.
4. **Calculate the x-coordinate of the vertex:**
$$ h = -\frac{-18}{2 \times 3} = \frac{18}{6} = 3 $$
5. **Calculate the y-coordinate of the vertex:**
$$ k = 3(3)^2 - 18(3) - 10 = 3 \times 9 - 54 - 10 = 27 - 54 - 10 = -37 $$
6. **Determine the direction of opening:** Since $$a = 3 > 0$$, the parabola opens **upward**.
7. **Conclusion:** The vertex is at $$ (3, -37) $$ and the parabola opens upward.
**Answer:** a.) Vertex: (3,-37) Opens: Upward