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

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1. **Write the equation and factor the coefficient of $$ x^2 $$:**  
   The quadratic function is  
   $$
   y = -2x^2 + 4x - 2.
   $$  
   Factor out $$-2$$ from the quadratic and linear terms:  
   $$
   y = -2\left(x^2 - 2x\right) - 2.
   $$

2. **Complete the square:**  
   To complete the square for $$ x^2 - 2x $$, take half the coefficient of $$ x $$ (which is $$-2$$) and square it:  
   $$
   \left(\frac{-2}{2}\right)^2 = 1.
   $$  
   Add and subtract $$1$$ inside the parentheses:  
   $$
   y = -2\left(x^2 - 2x + 1 - 1\right) - 2.
   $$  
   Thus,  
   $$
   y = -2\left[(x-1)^2 - 1\right] - 2.
   $$

3. **Rewrite in vertex form:**  
   Distribute the $$-2$$:  
   $$
   y = -2(x-1)^2 + 2 - 2.
   $$  
   Combine the constants:  
   $$
   y = -2(x-1)^2.
   $$  
   The vertex form is  
   $$
   y = -2(x-1)^2.
   $$

4. **Identify key features:**  
   - **Vertex:** The vertex of a quadratic in the form $$ y = a(x-h)^2 + k $$ is $$(h, k)$$. Here, the vertex is  
     $$
     (1, 0).
     $$
   - **Axis of symmetry:** This is the vertical line $$ x = h $$, so  
     $$
     x = 1.
     $$
   - **Direction of opening:** Since the coefficient $$ a = -2 $$ is negative, the parabola opens downward.
   - **$$ y $$-intercept:** Substitute $$ x = 0 $$ into the original equation:  
     $$
     y = -2(0)^2 + 4(0) - 2 = -2,
     $$  
     so the $$ y $$-intercept is  
     $$
     (0, -2).
     $$
   - **$$ x $$-intercepts:** Set $$ y = 0 $$:  
     $$
     0 = -2(x-1)^2.
     $$  
     Divide by $$-2$$:  
     $$
     0 = (x-1)^2.
     $$  
     Take the square root:  
     $$
     x - 1 = 0 \quad \Rightarrow \quad x = 1.
     $$  
     So there is a single $$ x $$-intercept at  
     $$
     (1, 0),
     $$  
     which is also the vertex.

5. **Sketching the graph:**  
   - Plot the vertex at $$(1, 0)$$.  
   - Draw the axis of symmetry as the vertical line $$ x = 1 $$.  
   - Mark the $$ y $$-intercept at $$(0, -2)$$.  
   - Note that the parabola touches the $$ x $$-axis only at the vertex $$(1, 0)$$ and opens downward.  
   - You can choose additional points for a smoother curve; for example, for $$ x = 2 $$:  
     $$
     y = -2(2-1)^2 = -2(1)^2 = -2,
     $$  
     which gives the point $$(2, -2)$$. Its symmetric point about $$ x=1 $$ is $$(0, -2)$$ (which is already plotted as the $$ y $$-intercept).

6. **Conclusion:**  
   The graph of  
   $$
   y = -2x^2 + 4x - 2
   $$  
   is a downward opening parabola with vertex $$(1, 0)$$, a single $$ x $$-intercept at $$(1, 0)$$, and a $$ y $$-intercept at $$(0, -2)$$.  
   
The key features are summarized as:  
- **Vertex:** $$ (1, 0) $$  
- **Axis of symmetry:** $$ x = 1 $$  
- **Opens:** Downward (since $$ a < 0 $$)  
- **$$ y $$-intercept:** $$ (0, -2) $$  
- **$$ x $$-intercept:** $$ (1, 0) $$

This completes the step-by-step solution for graphing the quadratic function.
To graph the quadratic function $$ y = -2x^2 + 4x - 2 $$, follow these steps: 1. **Identify the vertex form:** Rewrite the equation as $$ y = -2(x - 1)^2 $$. 2. **Determine key features:** - **Vertex:** $$(1, 0)$$ - **Axis of symmetry:** $$ x = 1 $$ - **Direction:** Opens downward - **Intercepts:** $$ y $$-intercept at $$(0, -2)$$ and $$ x $$-intercept at $$(1, 0)$$ 3. **Plot the graph:** - Draw the axis of symmetry at $$ x = 1 $$. - Plot the vertex at $$(1, 0)$$ and the $$ y $$-intercept at $$(0, -2)$$. - Since the parabola opens downward, sketch the curve symmetrically around the axis of symmetry. 4. **Conclusion:** The graph is a downward-opening parabola with vertex at $$(1, 0)$$, $$ y $$-intercept at $$(0, -2)$$, and $$ x $$-intercept at $$(1, 0)$$.

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

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