We are given the quadratic function $f(x) = -x^2 + 3x + 4$. Let's solve each part of the question:

1. The x-intercepts:

The x-intercepts are found by setting $f(x) = 0$ and solving for $x$. $0 = -x^2 + 3x + 4$ Rearrange the equation: $x^2 - 3x - 4 = 0$ Now, factor the quadratic equation: $(x - 4)(x + 1) = 0$ Thus, the solutions are: $x = 4 \quad \text{or} \quad x = -1$ So, the x-intercepts are $x = 4$ and $x = -1$.

2. The y-intercept:

The y-intercept occurs when $x = 0$. Plug $x = 0$ into the function: $f(0) = -(0)^2 + 3(0) + 4 = 4$ So, the y-intercept is $4$.

3. The vertex:

For a quadratic function of the form $f(x) = ax^2 + bx + c$, the x-coordinate of the vertex is given by: $x = -\frac{b}{2a}$ In this case, $a = -1$ and $b = 3$, so: $x = -\frac{3}{2(-1)} = \frac{3}{2}$ Now, substitute $x = \frac{3}{2}$ back into the function to find the y-coordinate: $f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) + 4$ $f\left(\frac{3}{2}\right) = -\frac{9}{4} + \frac{9}{2} + 4 = -\frac{9}{4} + \frac{18}{4} + \frac{16}{4} = \frac{25}{4}$ Thus, the vertex is $\left(\frac{3}{2}, \frac{25}{4}\right)$.

4. The line of symmetry:

The line of symmetry is the vertical line that passes through the vertex. Since the x-coordinate of the vertex is $\frac{3}{2}$, the equation of the line of symmetry is: $x = \frac{3}{2}$

Would you like further details or explanations on any part?

Here are five related questions to deepen understanding:

1. How do you factor a quadratic equation when it cannot be easily factored?
2. Can you find the vertex of any quadratic function by completing the square?
3. How does the sign of the coefficient $a$ affect the direction of a parabola?
4. What is the significance of the discriminant in finding the x-intercepts?
5. Can a quadratic function have no x-intercepts?

Tip: Always check if the quadratic equation can be factored easily before using the quadratic formula; it can save time!