Worksheet for graphing quadratic functions with six equations and corresponding coordinate grids.

To solve the problem of graphing quadratic functions, we need to follow a systematic approach. Let's break it down step by step using the general form of a quadratic function:

General Form of a Quadratic Function:

The standard form of a quadratic function is:
\[
f(x) = ax^2 + bx + c
\]
where \(a\), \(b\), and \(c\) are constants, and \(a \neq 0\).

Key Steps for Graphing a Quadratic Function:

1. Identify the Vertex:
- The vertex of a parabola given by \(f(x) = ax^2 + bx + c\) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
- Once you have the \(x\)-coordinate of the vertex, substitute it back into the function to find the \(y\)-coordinate.

2. Determine the Axis of Symmetry:
- The axis of symmetry is a vertical line that passes through the vertex. Its equation is:
\[
x = -\frac{b}{2a}
\]

3. Find the \(y\)-Intercept:
- The \(y\)-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the function to find \(f(0)\).

4. Find the \(x\)-Intercepts (if any):
- The \(x\)-intercepts occur when \(f(x) = 0\). Solve the quadratic equation \(ax^2 + bx + c = 0\) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
- If the discriminant (\(b^2 - 4ac\)) is positive, there are two real \(x\)-intercepts. If it is zero, there is one \(x\)-intercept (the vertex touches the \(x\)-axis). If it is negative, there are no real \(x\)-intercepts.

5. Plot Additional Points:
- Choose a few \(x\)-values on either side of the vertex and calculate the corresponding \(y\)-values to get more points on the parabola.

6. Sketch the Parabola:
- Use the vertex, intercepts, and additional points to sketch the parabola. Remember that the parabola opens upwards if \(a > 0\) and downwards if \(a < 0\).

Example Problem:

Let's solve one of the problems from the image you uploaded. Suppose we have the quadratic function:
\[
f(x) = 2x^2 - 4x + 1
\]

#### Step 1: Identify the Vertex
- The coefficients are \(a = 2\), \(b = -4\), and \(c = 1\).
- The \(x\)-coordinate of the vertex is:
\[
x = -\frac{b}{2a} = -\frac{-4}{2 \cdot 2} = \frac{4}{4} = 1
\]
- Substitute \(x = 1\) into the function to find the \(y\)-coordinate:
\[
f(1) = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1
\]
- So, the vertex is \((1, -1)\).

#### Step 2: Determine the Axis of Symmetry
- The axis of symmetry is:
\[
x = 1
\]

#### Step 3: Find the \(y\)-Intercept
- The \(y\)-intercept occurs when \(x = 0\):
\[
f(0) = 2(0)^2 - 4(0) + 1 = 1
\]
- So, the \(y\)-intercept is \((0, 1)\).

#### Step 4: Find the \(x\)-Intercepts
- Solve \(2x^2 - 4x + 1 = 0\) using the quadratic formula:
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = \frac{2 \pm \sqrt{2}}{2}
\]
- So, the \(x\)-intercepts are:
\[
x = \frac{2 + \sqrt{2}}{2} \quad \text{and} \quad x = \frac{2 - \sqrt{2}}{2}
\]

#### Step 5: Plot Additional Points
- Choose \(x = -1\):
\[
f(-1) = 2(-1)^2 - 4(-1) + 1 = 2 + 4 + 1 = 7
\]
Point: \((-1, 7)\)

- Choose \(x = 2\):
\[
f(2) = 2(2)^2 - 4(2) + 1 = 8 - 8 + 1 = 1
\]
Point: \((2, 1)\)

#### Step 6: Sketch the Parabola
- Plot the vertex \((1, -1)\), the \(y\)-intercept \((0, 1)\), the \(x\)-intercepts \(\left(\frac{2 + \sqrt{2}}{2}, 0\right)\) and \(\left(\frac{2 - \sqrt{2}}{2}, 0\right)\), and the additional points \((-1, 7)\) and \((2, 1)\).
- Draw a smooth parabola through these points, noting that it opens upwards because \(a = 2 > 0\).

Final Answer:

\[
\boxed{(1, -1)}
\]
This is the vertex of the parabola for the function \(f(x) = 2x^2 - 4x + 1\). You can follow similar steps for other quadratic functions in the worksheet.