The function presented in the image is:

$g(x) = -\frac{1}{4}x^2 - x + 4$

Let's break down the questions and solutions:

1. Vertex of the function: The vertex of a quadratic function $g(x) = ax^2 + bx + c$ can be found using the formula for the $x$-coordinate of the vertex:

   $x = -\frac{b}{2a}$

   Here, $a = -\frac{1}{4}$ and $b = -1$, so:

   $x = -\frac{-1}{2 \times -\frac{1}{4}} = -2$

   To find the $y$-coordinate of the vertex, substitute $x = -2$ into the original equation:

   $g(-2) = -\frac{1}{4}(-2)^2 - (-2) + 4 = -\frac{1}{4}(4) + 2 + 4 = -1 + 2 + 4 = 5$

   Therefore, the vertex is $(-2, 5)$, which matches the correct answer in the image.

2. Equation of the line of symmetry: The axis of symmetry is the vertical line that passes through the vertex. Since the $x$-coordinate of the vertex is $-2$, the equation of the line of symmetry is:

   $x = -2$

3. Maximum or minimum value: Since the leading coefficient $a = -\frac{1}{4}$ is negative, the parabola opens downward, meaning the function has a maximum value at the vertex. The maximum value is the $y$-coordinate of the vertex, which is 5.

4. The x-intercepts: To find the x-intercepts, set $g(x) = 0$ and solve for $x$:

   $-\frac{1}{4}x^2 - x + 4 = 0$

   Multiply through by 4 to eliminate the fraction:

   $-x^2 - 4x + 16 = 0$

   Multiply by $-1$ to simplify:

   $x^2 + 4x - 16 = 0$

   Now, solve this quadratic equation using the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   For $a = 1$, $b = 4$, and $c = -16$:

   $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-16)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 64}}{2} = \frac{-4 \pm \sqrt{80}}{2}$

   Simplifying:

   $x = \frac{-4 \pm 4\sqrt{5}}{2} = -2 \pm 2\sqrt{5}$

   This gives the x-intercepts as:

   $x_1 = -2 + 2\sqrt{5} \approx 2.47, \quad x_2 = -2 - 2\sqrt{5} \approx -6.47$

   Therefore, the correct x-intercepts are approximately $(2.47, 0)$ and $(-6.47, 0)$. It looks like you entered these values as $(2.46, 0)$ and $(-6.46, 0)$, which were marked incorrect likely because of rounding. Using more precise values would be correct.

5. The y-intercept: The y-intercept occurs when $x = 0$. Substituting into the equation for $g(x)$:

   $g(0) = -\frac{1}{4}(0)^2 - 0 + 4 = 4$

   Therefore, the y-intercept is $(0, 4)$.

Would you like further details or have any questions about this solution?

Here are 5 related questions to consider:

1. How do you determine whether a quadratic function has a maximum or minimum value?
2. How would the vertex change if the coefficient of $x^2$ were positive instead of negative?
3. Can the quadratic formula be used for any quadratic equation, and why?
4. How would you graph the function $g(x)$ based on the vertex, intercepts, and axis of symmetry?
5. What is the significance of the discriminant $b^2 - 4ac$ in the quadratic formula?

Tip: Always be mindful of rounding errors when dealing with approximate answers. When in doubt, retain as many decimal places as possible before submitting final results.