Worksheet for graphing quadratic functions, identifying vertex and axis of symmetry.

To solve the problem of graphing quadratic functions and identifying their key features (vertex, axis of symmetry, and intercepts), we will go through each function step by step. The general form of a quadratic function is:

$$
f(x) = ax^2 + bx + c
$$

Key Features of a Quadratic Function:

1. Vertex: The vertex of a parabola is given by the point \((h, k)\), where:
$$
h = -\frac{b}{2a}, \quad k = f(h)
$$
2. Axis of Symmetry: The axis of symmetry is a vertical line passing through the vertex, given by:
$$
x = -\frac{b}{2a}
$$
3. Intercepts:
- x-intercepts: Solve \( f(x) = 0 \) to find the roots.
- y-intercept: Evaluate \( f(0) \).

Let's solve each part of the problem.

---

Problem 1: \( g(x) = -2x^2 + 4x + 2 \)

#### Step 1: Identify the coefficients
Here, \( a = -2 \), \( b = 4 \), and \( c = 2 \).

#### Step 2: Find the vertex
The x-coordinate of the vertex is:
$$
h = -\frac{b}{2a} = -\frac{4}{2(-2)} = -\frac{4}{-4} = 1
$$
The y-coordinate of the vertex is:
$$
k = g(1) = -2(1)^2 + 4(1) + 2 = -2 + 4 + 2 = 4
$$
So, the vertex is:
$$
(1, 4)
$$

#### Step 3: Find the axis of symmetry
The axis of symmetry is:
$$
x = 1
$$

#### Step 4: Find the intercepts
- y-intercept: Set \( x = 0 \):
$$
g(0) = -2(0)^2 + 4(0) + 2 = 2
$$
So, the y-intercept is:
$$
(0, 2)
$$
- x-intercepts: Solve \( g(x) = 0 \):
$$
-2x^2 + 4x + 2 = 0
$$
Divide the equation by \(-2\):
$$
x^2 - 2x - 1 = 0
$$
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
$$
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}
$$
So, the x-intercepts are:
$$
(1 + \sqrt{2}, 0) \quad \text{and} \quad (1 - \sqrt{2}, 0)
$$

#### Final Answer for Problem 1:
- Vertex: \((1, 4)\)
- Axis of Symmetry: \( x = 1 \)
- y-intercept: \((0, 2)\)
- x-intercepts: \((1 + \sqrt{2}, 0)\) and \((1 - \sqrt{2}, 0)\)

---

Problem 2: \( h(x) = x^2 - x - 3 \)

#### Step 1: Identify the coefficients
Here, \( a = 1 \), \( b = -1 \), and \( c = -3 \).

#### Step 2: Find the vertex
The x-coordinate of the vertex is:
$$
h = -\frac{b}{2a} = -\frac{-1}{2(1)} = \frac{1}{2}
$$
The y-coordinate of the vertex is:
$$
k = h\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 3 = \frac{1}{4} - \frac{1}{2} - 3 = \frac{1}{4} - \frac{2}{4} - \frac{12}{4} = -\frac{13}{4}
$$
So, the vertex is:
$$
\left(\frac{1}{2}, -\frac{13}{4}\right)
$$

#### Step 3: Find the axis of symmetry
The axis of symmetry is:
$$
x = \frac{1}{2}
$$

#### Step 4: Find the intercepts
- y-intercept: Set \( x = 0 \):
$$
h(0) = 0^2 - 0 - 3 = -3
$$
So, the y-intercept is:
$$
(0, -3)
$$
- x-intercepts: Solve \( h(x) = 0 \):
$$
x^2 - x - 3 = 0
$$
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
$$
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}
$$
So, the x-intercepts are:
$$
\left(\frac{1 + \sqrt{13}}{2}, 0\right) \quad \text{and} \quad \left(\frac{1 - \sqrt{13}}{2}, 0\right)
$$

#### Final Answer for Problem 2:
- Vertex: \(\left(\frac{1}{2}, -\frac{13}{4}\right)\)
- Axis of Symmetry: \( x = \frac{1}{2} \)
- y-intercept: \((0, -3)\)
- x-intercepts: \(\left(\frac{1 + \sqrt{13}}{2}, 0\right)\) and \(\left(\frac{1 - \sqrt{13}}{2}, 0\right)\)

---

Problem 3: \( k(x) = -x^2 + 2x + 3 \)

#### Step 1: Identify the coefficients
Here, \( a = -1 \), \( b = 2 \), and \( c = 3 \).

#### Step 2: Find the vertex
The x-coordinate of the vertex is:
$$
h = -\frac{b}{2a} = -\frac{2}{2(-1)} = -\frac{2}{-2} = 1
$$
The y-coordinate of the vertex is:
$$
k = k(1) = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4
$$
So, the vertex is:
$$
(1, 4)
$$

#### Step 3: Find the axis of symmetry
The axis of symmetry is:
$$
x = 1
$$

#### Step 4: Find the intercepts
- y-intercept: Set \( x = 0 \):
$$
k(0) = -(0)^2 + 2(0) + 3 = 3
$$
So, the y-intercept is:
$$
(0, 3)
$$
- x-intercepts: Solve \( k(x) = 0 \):
$$
-x^2 + 2x + 3 = 0
$$
Multiply by \(-1\):
$$
x^2 - 2x - 3 = 0
$$
Factorize:
$$
(x - 3)(x + 1) = 0
$$
So, the x-intercepts are:
$$
(3, 0) \quad \text{and} \quad (-1, 0)
$$

#### Final Answer for Problem 3:
- Vertex: \((1, 4)\)
- Axis of Symmetry: \( x = 1 \)
- y-intercept: \((0, 3)\)
- x-intercepts: \((3, 0)\) and \((-1, 0)\)

---

Final Answers

1. \( g(x) = -2x^2 + 4x + 2 \):
- Vertex: \((1, 4)\)
- Axis of Symmetry: \( x = 1 \)
- y-intercept: \((0, 2)\)
- x-intercepts: \((1 + \sqrt{2}, 0)\) and \((1 - \sqrt{2}, 0)\)

2. \( h(x) = x^2 - x - 3 \):
- Vertex: \(\left(\frac{1}{2}, -\frac{13}{4}\right)\)
- Axis of Symmetry: \( x = \frac{1}{2} \)
- y-intercept: \((0, -3)\)
- x-intercepts: \(\left(\frac{1 + \sqrt{13}}{2}, 0\right)\) and \(\left(\frac{1 - \sqrt{13}}{2}, 0\right)\)

3. \( k(x) = -x^2 + 2x + 3 \):
- Vertex: \((1, 4)\)
- Axis of Symmetry: \( x = 1 \)
- y-intercept: \((0, 3)\)
- x-intercepts: \((3, 0)\) and \((-1, 0)\)

$$
\boxed{
\begin{array}{l}
\text{1. Vertex: } (1, 4), \text{ Axis of Symmetry: } x = 1, \text{ y-intercept: } (0, 2), \text{ x-intercepts: } (1 + \sqrt{2}, 0) \text{ and } (1 - \sqrt{2}, 0) \\
\text{2. Vertex: } \left(\frac{1}{2}, -\frac{13}{4}\right), \text{ Axis of Symmetry: } x = \frac{1}{2}, \text{ y-intercept: } (0, -3), \text{ x-intercepts: } \left(\frac{1 + \sqrt{13}}{2}, 0\right) \text{ and } \left(\frac{1 - \sqrt{13}}{2}, 0\right) \\
\text{3. Vertex: } (1, 4), \text{ Axis of Symmetry: } x = 1, \text{ y-intercept: } (0, 3), \text{ x-intercepts: } (3, 0) \text{ and } (-1, 0)
\end{array}
}
$$