Consider the following function. Complete parts (a) through (e) below. $$ y=-2 x^{2}+8 x+4 $$ The parabola opens upward. The parabola opens downward. b. Find the vertex. The vertex is $$ \square $$. (Type an ordered pair.) c. Use the quadratic formula to find the x-intercepts. The x-intercept(s) is/are $$ \square $$. (Type an integer or a decimal rounded to the nearest tenth as needed. Use a comma to separate answers as needed.) d. Find the $$ y $$-intercept. The $$ y $$-intercept is $$ \square $$. (Type an integer or a fraction.) e. Use the results from parts (a)-(d) to graph the quadratic . .

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**a. Orientation of the Parabola**

The function is  
$$
y=-2x^2+8x+4.
$$  
Since the coefficient of $$ x^2 $$ is $$-2$$ (which is negative), the parabola opens **downward**.

---

**b. Vertex**

The $$ x $$-coordinate of the vertex is given by  
$$
x=-\frac{b}{2a}.
$$  
Here, $$ a=-2 $$ and $$ b=8 $$, so  
$$
x=-\frac{8}{2(-2)}=-\frac{8}{-4}=2.
$$

Now, substitute $$ x=2 $$ into the function to find the $$ y $$-coordinate:  
$$
y=-2(2)^2+8(2)+4=-2(4)+16+4=-8+16+4=12.
$$

Thus, the vertex is  
$$
(2,\,12).
$$

---

**c. $$ x $$-Intercepts**

To find the $$ x $$-intercepts, set $$ y=0 $$:
$$
-2x^2+8x+4=0.
$$

It is often easier to work with a positive leading coefficient, so multiply both sides by $$-1$$:
$$
2x^2-8x-4=0.
$$

Divide by $$2$$ to simplify:
$$
x^2-4x-2=0.
$$

Now, apply the quadratic formula  
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
$$
with $$ a=1 $$, $$ b=-4 $$, and $$ c=-2 $$:
  
Calculate the discriminant:
$$
b^2-4ac=(-4)^2-4(1)(-2)=16+8=24.
$$

Thus,
$$
x=\frac{-(-4)\pm\sqrt{24}}{2(1)}=\frac{4\pm2\sqrt{6}}{2}=2\pm\sqrt{6}.
$$

So the $$ x $$-intercepts are  
$$
(2-\sqrt{6},\,0) \quad \text{and} \quad (2+\sqrt{6},\,0).
$$

Rounded to the nearest tenth, $$ \sqrt{6}\approx2.4 $$, so approximately:
$$
(2-2.4,\,0)=(-0.4,\,0) \quad \text{and} \quad (2+2.4,\,0)=(4.4,\,0).
$$

---

**d. $$ y $$-Intercept**

The $$ y $$-intercept is found by setting $$ x=0 $$:
$$
y=-2(0)^2+8(0)+4=4.
$$

Thus, the $$ y $$-intercept is  
$$
(0,\,4).
$$

---

**e. Graphing the Quadratic Function**

To graph the function $$ y=-2x^2+8x+4 $$, plot the following key features:

- **Vertex:** $$ (2,\,12) $$.  
- **Axis of Symmetry:** The line $$ x=2 $$.
- **$$ x $$-Intercepts:** $$ (2-\sqrt{6},\,0) $$ and $$ (2+\sqrt{6},\,0) $$ (approximately $$(-0.4,\,0)$$ and $$(4.4,\,0)$$).
- **$$ y $$-Intercept:** $$ (0,\,4) $$.

Since the parabola opens downward (because the coefficient of $$ x^2 $$ is $$-2$$), sketch a downward-opening curve using these points. The vertex is the highest point on the graph, and the curve is symmetric about the line $$ x=2 $$.

**a. Orientation of the Parabola** The parabola opens **downward**. --- **b. Vertex** The vertex is $$ (2,\,12). $$ --- **c. $$ x $$-Intercepts** The $$ x $$-intercepts are $$ (-0.4,\,0) \quad \text{and} \quad (4.4,\,0). $$ --- **d. $$ y $$-Intercept** The $$ y $$-intercept is $$ (0,\,4). $$ --- **e. Graphing the Quadratic Function** To graph $$ y=-2x^2+8x+4 $$, plot the vertex at $$ (2,\,12) $$, the $$ y $$-intercept at $$ (0,\,4) $$, and the $$ x $$-intercepts at approximately $$ (-0.4,\,0) $$ and $$ (4.4,\,0) $$. Draw a downward-opening parabola symmetric about the line $$ x=2 $$.

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