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

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**Step 1. Rewrite the Function**

We are given the quadratic function
$$
y = -x^{2} + 2x + 4.
$$

**Step 2. Find the Vertex**

For a quadratic function in the form
$$
y = ax^{2}+bx+c,
$$
the $$x$$-coordinate of the vertex is found by
$$
x = -\frac{b}{2a}.
$$
Here, $$a = -1$$ and $$b = 2$$. Thus,
$$
x = -\frac{2}{2(-1)} = -\frac{2}{-2} = 1.
$$

Now substitute $$x = 1$$ into the function to find the $$y$$-coordinate:
$$
y = - (1)^{2} + 2(1) + 4 = -1 + 2 + 4 = 5.
$$
So, the vertex is at
$$
(1, 5).
$$

**Step 3. Find the $$x$$-Intercepts**

Set $$y = 0$$ to solve for $$x$$:
$$
0 = -x^{2} + 2x + 4.
$$
Multiply both sides by $$-1$$ to simplify:
$$
0 = x^{2} - 2x - 4.
$$

Apply the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a},
$$
with $$a = 1$$, $$b = -2$$, and $$c = -4$$:
$$
x = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(-4)}}{2(1)} 
= \frac{2 \pm \sqrt{4 + 16}}{2} 
= \frac{2 \pm \sqrt{20}}{2}.
$$

Simplify $$\sqrt{20}$$:
$$
\sqrt{20} = 2\sqrt{5}.
$$

Thus,
$$
x = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}.
$$
The $$x$$-intercepts are:
$$
x = 1 + \sqrt{5} \quad \text{and} \quad x = 1 - \sqrt{5}.
$$
Approximately,
$$
1 + \sqrt{5} \approx 3.236 \quad \text{and} \quad 1 - \sqrt{5} \approx -1.236.
$$

**Step 4. Find the $$y$$-Intercept**

Let $$x=0$$:
$$
y = - (0)^{2} + 2(0) + 4 = 4.
$$
So, the $$y$$-intercept is at
$$
(0, 4).
$$

**Step 5. Identify the Opening Direction and Axis of Symmetry**

Since the coefficient of $$x^2$$ is negative ($$a=-1$$), the parabola opens downward.  
The axis of symmetry is the vertical line through the vertex:
$$
x = 1.
$$

**Step 6. Sketching the Graph**

1. Plot the vertex at $$(1, 5)$$.
2. Plot the $$y$$-intercept at $$(0, 4)$$.
3. Plot the $$x$$-intercepts at approximately $$(-1.236, 0)$$ and $$(3.236, 0)$$.
4. Draw the axis of symmetry $$x = 1$$.
5. Sketch a smooth, downward-opening parabola passing through these points.

The graph should look like a downward opening parabola with its highest point at $$(1, 5)$$.
To graph the quadratic function $$ y = -x^{2} + 2x + 4 $$: 1. **Vertex**: The vertex is at $$(1, 5)$$. 2. **$$x$$-Intercepts**: Approximately at $$(-1.236, 0)$$ and $$(3.236, 0)$$. 3. **$$y$$-Intercept**: At $$(0, 4)$$. 4. **Axis of Symmetry**: The vertical line $$x = 1$$. 5. **Parabola**: Opens downward. Plot these points and sketch the downward-opening parabola accordingly.

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

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