Graph the function. 3. $$ y=-3 x^{2}+6 x-4 $$

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**Step 1. Write the function and identify the coefficients**

We are given
$$
y=-3x^2+6x-4.
$$
Here, the quadratic coefficient is $$ a=-3 $$, the linear coefficient is $$ b=6 $$, and the constant term is $$ c=-4 $$.

---

**Step 2. Convert to Vertex Form**

The vertex form of a quadratic is
$$
y=a(x-h)^2+k,
$$
where $$(h,k)$$ is the vertex.

Start by factoring out $$-3$$ from the first two terms:
$$
y=-3\left(x^2-2x\right)-4.
$$

Complete the square inside the parentheses. Notice
$$
x^2-2x = \left(x-1\right)^2-1,
$$
so substitute:
$$
y=-3\left[\left(x-1\right)^2-1\right]-4.
$$

Distribute the $$-3$$:
$$
y=-3\left(x-1\right)^2+3-4,
$$
which simplifies to
$$
y=-3\left(x-1\right)^2-1.
$$

Thus, the vertex form is
$$
y=-3\left(x-1\right)^2-1.
$$

---

**Step 3. Identify Key Features**

1. **Vertex:** From the vertex form, the vertex is
   $$
   (h,k)=(1,-1).
   $$

2. **Axis of Symmetry:** The vertical line through the vertex is 
   $$
   x=1.
   $$

3. **Direction of Opening:** Since $$ a=-3 $$ is negative, the parabola opens downward.

4. **Y-Intercept:** Set $$x=0$$ in the original function:
   $$
   y=-3(0)^2+6(0)-4=-4.
   $$
   The y-intercept is $$(0,-4)$$.

5. **X-Intercepts:** Solve
   $$
   -3x^2+6x-4=0.
   $$
   Multiply both sides by $$-1$$:
   $$
   3x^2-6x+4=0.
   $$
   Apply the quadratic formula:
   $$
   x=\frac{6\pm\sqrt{(-6)^2-4\cdot3\cdot4}}{2\cdot3}
   =\frac{6\pm\sqrt{36-48}}{6}
   =\frac{6\pm\sqrt{-12}}{6}.
   $$
   Since the discriminant is negative, there are no real x-intercepts.

6. **Symmetric Point:** The y-intercept $$(0,-4)$$ is on the left of the axis. Its symmetric point with respect to the axis $$x=1$$ is found by reflecting $$x=0$$ to $$x=2$$ (since $$1-0=1$$ then $$1+1=2$$). Therefore, the point $$(2,-4)$$ is also on the graph.

---

**Step 4. Sketching the Graph**

- Plot the vertex at $$(1,-1)$$.
- Draw the axis of symmetry, the vertical line $$x=1$$.
- Plot the y-intercept $$(0,-4)$$ and its symmetric point $$(2,-4)$$.
- Note that the parabola opens downward, so it reaches a maximum at the vertex.
- Since there are no real x-intercepts, the graph does not cross the x-axis.

---

The resulting graph is a downward-opening parabola with vertex $$(1,-1)$$, symmetric about $$x=1$$, passing through $$(0,-4)$$ and $$(2,-4)$$.
To graph the function $$ y = -3x^2 + 6x - 4 $$: 1. **Vertex**: The vertex is at $$(1, -1)$$. 2. **Axis of Symmetry**: $$ x = 1 $$. 3. **Y-Intercept**: $$(0, -4)$$. 4. **No X-Intercepts**. 5. **Symmetric Point**: $$(2, -4)$$. The parabola opens downward, with its highest point at the vertex. It passes through the points $$(0, -4)$$ and $$(2, -4)$$.

Graph the function. 3. \( y=-3 x^{2}+6 x-4 \)

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