Graph the quadratic function below. $$ y=-2 x^{2}-4 x-2 $$

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**Step 1. Identify the quadratic equation and its coefficients**

We are given the quadratic function  
$$
y = -2x^2 - 4x - 2.
$$  
Here,  
$$
a = -2,\quad b = -4,\quad c = -2.
$$

**Step 2. Find the vertex**

The $$x$$-coordinate of the vertex for a quadratic $$ax^2+bx+c$$ is given by  
$$
x_v = -\frac{b}{2a}.
$$  
Substitute the values:  
$$
x_v = -\frac{-4}{2(-2)} = \frac{4}{-4} = -1.
$$  
Now, substitute $$x_v = -1$$ into the function to get the $$y$$-coordinate:  
$$
y_v = -2(-1)^2 - 4(-1) - 2 = -2(1) + 4 - 2 = 0.
$$  
Thus, the vertex is at  
$$
(-1,\, 0).
$$

**Step 3. Determine the axis of symmetry**

The axis of symmetry is the vertical line through the vertex:  
$$
x = -1.
$$

**Step 4. Determine the direction of opening**

Since $$a = -2 < 0$$, the parabola opens downward.

**Step 5. Find the $$y$$-intercept**

Set $$x = 0$$:  
$$
y = -2(0)^2 - 4(0) - 2 = -2.
$$  
Thus, the $$y$$-intercept is at  
$$
(0,\, -2).
$$

**Step 6. Find the $$x$$-intercepts**

Set $$y = 0$$:  
$$
-2x^2 - 4x - 2 = 0.
$$  
Divide the entire equation by $$-2$$:  
$$
x^2 + 2x + 1 = 0.
$$  
Factor the quadratic:  
$$
(x + 1)^2 = 0.
$$  
Hence, the only solution is $$x = -1$$. This means the $$x$$-intercept is at  
$$
(-1,\, 0),
$$  
which coincides with the vertex.

**Step 7. Find additional points**

To sketch the graph, we can find points equidistant from the axis of symmetry.

- Choose $$x = -2$$ (which is $$1$$ unit left of the vertex):  
  $$
  y = -2(-2)^2 - 4(-2) - 2 = -2(4) + 8 - 2 = -8 + 8 - 2 = -2.
  $$  
  So, one point is $$(-2,\, -2)$$.

- By symmetry, the corresponding point on the right side is $$x = 0$$ (which is $$1$$ unit right of the vertex):  
  We already found that $$(0,\, -2)$$ is the $$y$$-intercept.

- Choose $$x = 1$$ (which is $$2$$ units right of the vertex):  
  $$
  y = -2(1)^2 - 4(1) - 2 = -2 - 4 - 2 = -8.
  $$
  By symmetry, the corresponding point is at $$x = -3$$:  
  $$
  y = -2(-3)^2 - 4(-3) - 2 = -2(9) + 12 - 2 = -18 + 12 - 2 = -8.
  $$
  So, we have points $$(1,\, -8)$$ and $$(-3,\, -8)$$.

**Step 8. Sketching the graph**

Using the points found:

- **Vertex:** $$(-1,\, 0)$$  
- **Axis of symmetry:** $$x = -1$$  
- **$$y$$-intercept:** $$(0,\, -2)$$  
- **Additional points:** $$(-2,\, -2)$$, $$(-3,\, -8)$$, and $$(1,\, -8)$$  
- **$$x$$-intercept:** $$(-1,\, 0)$$ (a double root)

The parabola opens downward, with the highest point at the vertex $$(-1, 0)$$.

Plot these points on the coordinate plane, draw a smooth curve through them ensuring symmetry about the line $$x = -1$$, and label the vertex and intercepts accordingly.
To graph the quadratic function $$ y = -2x^2 - 4x - 2 $$: 1. **Vertex**: $$(-1, 0)$$ 2. **Axis of Symmetry**: $$x = -1$$ 3. **Direction**: Opens downward 4. **Intercepts**: - $$y$$-intercept: $$(0, -2)$$ - $$x$$-intercept: $$(-1, 0)$$ 5. **Additional Points**: - $$(-2, -2)$$ - $$(-3, -8)$$ - $$(1, -8)$$ Plot these points and draw a downward-opening parabola symmetric about $$x = -1$$.

Graph the quadratic function below. \( y=-2 x^{2}-4 x-2 \)

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