Math worksheet for graphing quadratic functions from standard form, including finding vertex, axis of symmetry, intercepts, and more.

Problem: Analyze and graph the given quadratic functions.

We are tasked with analyzing two quadratic functions:
1. \( f(x) = 3x^2 \)
2. \( f(x) = x^2 + 2x + 1 \)

For each function, we need to find:
- Vertex
- Axis of symmetry
- x-intercepts
- y-intercept
- Maximum/Minimum value
- Domain
- Range

Finally, we will graph the parabolas.

---

Function 1: \( f(x) = 3x^2 \)

#### 1. Vertex
The general form of a quadratic function is \( f(x) = ax^2 + bx + c \). For \( f(x) = 3x^2 \):
- \( a = 3 \), \( b = 0 \), \( c = 0 \).

The vertex of a parabola in standard form is given by:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = 3 \) and \( b = 0 \):
\[
x = -\frac{0}{2 \cdot 3} = 0
\]
To find the \( y \)-coordinate of the vertex, substitute \( x = 0 \) into \( f(x) \):
\[
f(0) = 3(0)^2 = 0
\]
Thus, the vertex is:
\[
\boxed{(0, 0)}
\]

#### 2. Axis of Symmetry
The axis of symmetry is a vertical line passing through the vertex. Since the vertex is at \( x = 0 \), the axis of symmetry is:
\[
x = 0
\]

#### 3. x-intercepts
The x-intercepts occur where \( f(x) = 0 \):
\[
3x^2 = 0
\]
Solve for \( x \):
\[
x^2 = 0 \implies x = 0
\]
Thus, the x-intercept is:
\[
\boxed{(0, 0)}
\]

#### 4. y-intercept
The y-intercept occurs where \( x = 0 \):
\[
f(0) = 3(0)^2 = 0
\]
Thus, the y-intercept is:
\[
\boxed{(0, 0)}
\]

#### 5. Maximum/Minimum Value
Since \( a = 3 > 0 \), the parabola opens upwards, and the vertex represents the minimum value. The minimum value is:
\[
f(0) = 0
\]

#### 6. Domain
The domain of a quadratic function is all real numbers:
\[
\boxed{(-\infty, \infty)}
\]

#### 7. Range
Since the parabola opens upwards and the minimum value is 0, the range is:
\[
\boxed{[0, \infty)}
\]

#### 8. Graph
The parabola \( f(x) = 3x^2 \) is a standard parabola that opens upwards, with its vertex at the origin. It is narrower than the parabola \( f(x) = x^2 \) because the coefficient of \( x^2 \) is 3.

---

Function 2: \( f(x) = x^2 + 2x + 1 \)

#### 1. Vertex
For \( f(x) = x^2 + 2x + 1 \):
- \( a = 1 \), \( b = 2 \), \( c = 1 \).

The vertex is given by:
\[
x = -\frac{b}{2a}
\]
Substitute \( a = 1 \) and \( b = 2 \):
\[
x = -\frac{2}{2 \cdot 1} = -1
\]
To find the \( y \)-coordinate of the vertex, substitute \( x = -1 \) into \( f(x) \):
\[
f(-1) = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0
\]
Thus, the vertex is:
\[
\boxed{(-1, 0)}
\]

#### 2. Axis of Symmetry
The axis of symmetry is a vertical line passing through the vertex. Since the vertex is at \( x = -1 \), the axis of symmetry is:
\[
x = -1
\]

#### 3. x-intercepts
The x-intercepts occur where \( f(x) = 0 \):
\[
x^2 + 2x + 1 = 0
\]
Factor the quadratic:
\[
(x + 1)^2 = 0
\]
Solve for \( x \):
\[
x + 1 = 0 \implies x = -1
\]
Thus, the x-intercept is:
\[
\boxed{(-1, 0)}
\]

#### 4. y-intercept
The y-intercept occurs where \( x = 0 \):
\[
f(0) = 0^2 + 2(0) + 1 = 1
\]
Thus, the y-intercept is:
\[
\boxed{(0, 1)}
\]

#### 5. Maximum/Minimum Value
Since \( a = 1 > 0 \), the parabola opens upwards, and the vertex represents the minimum value. The minimum value is:
\[
f(-1) = 0
\]

#### 6. Domain
The domain of a quadratic function is all real numbers:
\[
\boxed{(-\infty, \infty)}
\]

#### 7. Range
Since the parabola opens upwards and the minimum value is 0, the range is:
\[
\boxed{[0, \infty)}
\]

#### 8. Graph
The parabola \( f(x) = x^2 + 2x + 1 \) is a standard parabola that opens upwards, with its vertex at \( (-1, 0) \). It is the same shape as \( f(x) = x^2 \) but shifted to the left by 1 unit.

---

Final Answers

#### Function 1: \( f(x) = 3x^2 \)
- Vertex: \( (0, 0) \)
- Axis of symmetry: \( x = 0 \)
- x-intercept: \( (0, 0) \)
- y-intercept: \( (0, 0) \)
- Minimum value: \( 0 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [0, \infty) \)

#### Function 2: \( f(x) = x^2 + 2x + 1 \)
- Vertex: \( (-1, 0) \)
- Axis of symmetry: \( x = -1 \)
- x-intercept: \( (-1, 0) \)
- y-intercept: \( (0, 1) \)
- Minimum value: \( 0 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [0, \infty) \)

---

Graphs

1. For \( f(x) = 3x^2 \):
- Vertex at \( (0, 0) \)
- Opens upwards
- Narrower than \( f(x) = x^2 \)

2. For \( f(x) = x^2 + 2x + 1 \):
- Vertex at \( (-1, 0) \)
- Opens upwards
- Same shape as \( f(x) = x^2 \) but shifted left by 1 unit

Final Answer:
\[
\boxed{
\begin{array}{l|l}
\text{Function 1: } f(x) = 3x^2 & \text{Function 2: } f(x) = x^2 + 2x + 1 \\
\hline
\text{Vertex: } (0, 0) & \text{Vertex: } (-1, 0) \\
\text{Axis of symmetry: } x = 0 & \text{Axis of symmetry: } x = -1 \\
\text{x-intercept: } (0, 0) & \text{x-intercept: } (-1, 0) \\
\text{y-intercept: } (0, 0) & \text{y-intercept: } (0, 1) \\
\text{Minimum value: } 0 & \text{Minimum value: } 0 \\
\text{Domain: } (-\infty, \infty) & \text{Domain: } (-\infty, \infty) \\
\text{Range: } [0, \infty) & \text{Range: } [0, \infty) \\
\end{array}
}
\]