$$ 1 \leftarrow \quad \begin{array}{l} ext { Sketch a graph of } f . \ f(x)=-3 x^{2}-2\end{array} $$

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$$ 1 \leftarrow \quad \begin{array}{l}	ext { Sketch a graph of } f . \ f(x)=-3 x^{2}-2\end{array} $$

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1. **Identify the function type and coefficients**  
   The given function is  
   $$
   f(x) = -3x^2 - 2.
   $$  
   This is a quadratic function in the form  
   $$
   f(x) = ax^2 + bx + c,
   $$  
   with  
   $$
   a = -3,\quad b = 0,\quad c = -2.
   $$  
   Since $$ a < 0 $$, the parabola opens downward.

2. **Determine the vertex**  
   The vertex of a quadratic function is found using  
   $$
   x = -\frac{b}{2a}.
   $$  
   Substituting $$ b = 0 $$ and $$ a = -3 $$:  
   $$
   x = -\frac{0}{2(-3)} = 0.
   $$  
   To find the $$ y $$-coordinate of the vertex, substitute $$ x = 0 $$ into $$ f(x) $$:  
   $$
   f(0) = -3(0)^2 - 2 = -2.
   $$  
   Thus, the vertex is at  
   $$
   (0, -2).
   $$

3. **Find the $$ y $$-intercept**  
   The $$ y $$-intercept occurs when $$ x = 0 $$. We already have  
   $$
   f(0) = -2.
   $$  
   So, the $$ y $$-intercept is  
   $$
   (0, -2).
   $$

4. **Determine the $$ x $$-intercepts**  
   To find the $$ x $$-intercepts, set $$ f(x) = 0 $$:  
   $$
   -3x^2 - 2 = 0.
   $$  
   Rearranging gives  
   $$
   -3x^2 = 2 \quad \Rightarrow \quad x^2 = -\frac{2}{3}.
   $$  
   Since the square of a real number cannot be negative, there are no real $$ x $$-intercepts.

5. **Identify the axis of symmetry**  
   For a quadratic function in the form $$ f(x) = ax^2 + bx + c $$, the axis of symmetry is  
   $$
   x = -\frac{b}{2a}.
   $$  
   With $$ b = 0 $$ and $$ a = -3 $$, the axis of symmetry is  
   $$
   x = 0.
   $$

6. **Plot additional points for clarity**  
   Choose points on either side of the vertex:
   - For $$ x = 1 $$:
     $$
     f(1) = -3(1)^2 - 2 = -3 - 2 = -5.
     $$
     So, one point is $$ (1, -5) $$.
   - For $$ x = -1 $$:
     $$
     f(-1) = -3(-1)^2 - 2 = -3 - 2 = -5.
     $$
     So, another point is $$ (-1, -5) $$.

7. **Sketch the graph**  
   - Plot the vertex $$ (0, -2) $$.
   - Plot the points $$ (1, -5) $$ and $$ (-1, -5) $$.
   - Draw the axis of symmetry along $$ x = 0 $$.
   - Since the parabola opens downward, sketch a smooth curve passing through these points, noting that there are no real $$ x $$-intercepts.

The resulting graph is a downward opening parabola with vertex at $$ (0, -2) $$ and symmetric points $$ (1, -5) $$ and $$ (-1, -5) $$.
To sketch the graph of $$ f(x) = -3x^2 - 2 $$: 1. **Vertex**: At $$ (0, -2) $$. 2. **Axis of Symmetry**: $$ x = 0 $$. 3. **No $$ x $$-intercepts**. 4. **Additional Points**: $$ (1, -5) $$ and $$ (-1, -5) $$. Draw a downward-opening parabola passing through these points.

\( 1 \leftarrow \quad \begin{array}{l}\text { Sketch a graph of } f . \\ f(x)=-3 x^{2}-2\end{array} \)

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