To analyze the function $y = -3x^2 + 6$ and answer the given statements, let's evaluate each one step by step.

Function Characteristics

1. Equation: $y = -3x^2 + 6$

   • This is a downward-facing parabola because the coefficient of $x^2$ is negative.

2. Vertex:

   • The vertex form of a parabola is $y = a(x-h)^2 + k$. Here, $h = 0$, $k = 6$, so the vertex is at $(0, 6)$.

3. Axis of Symmetry:

   • The axis of symmetry for a parabola is $x = h$, so here it is $x = 0$, not $y = 0$.

4. Range:

   • The maximum value of $y$ is at the vertex ($y = 6$), and since the parabola opens downward, the range is $y \leq 6$.

5. Quadrants:

   • The parabola exists in the 2nd and 4th quadrants because it passes through $y = 6$ at $x = 0$, and as $x \to \pm \infty$, $y \to -\infty$.

6. Point Check:

   • For the point $(-1, -3)$, substitute $x = -1$ into the equation: $y = -3(-1)^2 + 6 = -3 + 6 = 3$ Thus, $(-1, -3)$ does not lie on the parabola.

Verifying Statements:

• a. "It only exists in 3rd and 4th quadrants." — False. The parabola exists in the 2nd and 4th quadrants.
• b. "Range is in the set $\{ y \mid y \leq 6 \}$." — True. The range is $y \leq 6$.
• c. "Vertex is $(0, 6)$." — True.
• d. "Axis of symmetry is $y = 0$." — False. The axis of symmetry is $x = 0$.
• e. "Point $(-1, -3)$ exists in the parabola." — False. This point is not on the parabola.

Do you want further details or have questions about these explanations? Here are 5 related questions for deeper understanding:

1. What happens to the range if the coefficient of $x^2$ is positive instead of negative?
2. How can we rewrite the equation in vertex form?
3. What is the domain of this parabola?
4. Can we determine where the parabola intersects the x-axis or y-axis?
5. How do changes to the constant term affect the vertex?

Tip: Always start by analyzing the vertex and the sign of the $x^2$-coefficient to understand a parabola's orientation and range.