The image shows a quadratic function problem with the equation $f(x) = -3x^2 + 7x + 5$. The task asks for the following:

1. Find the vertex of the quadratic function (already partially calculated as $\left( \frac{7}{6}, \frac{109}{12} \right)$).
2. Determine whether the graph opens up or down.

Answer Explanation:

1. Vertex: The vertex form of a quadratic function $f(x) = ax^2 + bx + c$ is given by: $x = -\frac{b}{2a}$ For $f(x) = -3x^2 + 7x + 5$:

   • $a = -3$
   • $b = 7$

   Substituting into the vertex formula: $x = -\frac{7}{2(-3)} = \frac{7}{6}$ The $y$-coordinate of the vertex can be found by substituting $x = \frac{7}{6}$ back into the original equation: $f\left(\frac{7}{6}\right) = -3\left(\frac{7}{6}\right)^2 + 7\left(\frac{7}{6}\right) + 5 = \frac{109}{12}$ So, the vertex is $\left( \frac{7}{6}, \frac{109}{12} \right)$.

2. Graph opens up or down: The graph of a quadratic function opens up if $a > 0$ and opens down if $a < 0$. Here, $a = -3$, which is less than 0, so the graph opens down.

Would you like a more detailed explanation on any step or further assistance with the problem?

Related Questions:

1. How do you convert a quadratic equation into vertex form?
2. How do you find the axis of symmetry for a quadratic function?
3. What is the significance of the sign of $a$ in determining the direction of a parabola?
4. How do you calculate the y-intercept of a quadratic equation?
5. Can a quadratic function have more than one vertex?

Tip:

Always check the sign of the coefficient $a$ to quickly determine whether the parabola opens up or down.