I’m a high school math student with a C in Algebra. Explain the steps for how to graph y=2x^2-2x?

Graphing the quadratic function \( y = 2x^2 - 2x \) involves a few key steps. Here’s a step-by-step guide to help you understand the process:

### 1. **Identify the type of function:**
   - The equation \( y = 2x^2 - 2x \) is a **quadratic function**, which means its graph will be a **parabola**. Since the coefficient of \( x^2 \) (which is 2) is positive, the parabola will open upwards.

### 2. **Find the vertex:**
   - The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).
   - Here, \( a = 2 \) and \( b = -2 \).
   - Substituting the values:
     \[
     x = -\frac{-2}{2(2)} = \frac{2}{4} = \frac{1}{2}
     \]
   - Now, substitute \( x = \frac{1}{2} \) back into the equation to find the \( y \)-coordinate of the vertex:
     \[
     y = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) = 2\left(\frac{1}{4}\right) - 1 = \frac{1}{2} - 1 = -\frac{1}{2}
     \]
   - So, the vertex is \( \left(\frac{1}{2}, -\frac{1}{2}\right) \).

### 3. **Find the y-intercept:**
   - The \( y \)-intercept occurs when \( x = 0 \).
   - Substitute \( x = 0 \) into the equation:
     \[
     y = 2(0)^2 - 2(0) = 0
     \]
   - So, the \( y \)-intercept is \( (0, 0) \).

### 4. **Find the x-intercepts:**
   - The \( x \)-intercepts occur where \( y = 0 \).
   - Set the equation equal to zero and solve for \( x \):
     \[
     2x^2 - 2x = 0
     \]
     \[
     2x(x - 1) = 0
     \]
   - This gives two solutions: \( x = 0 \) and \( x = 1 \).
   - So, the \( x \)-intercepts are \( (0, 0) \) and \( (1, 0) \).

### 5. **Plot the points:**
   - Now that you have the vertex \( \left(\frac{1}{2}, -\frac{1}{2}\right) \), the y-intercept \( (0, 0) \), and the x-intercepts \( (0, 0) \) and \( (1, 0) \), you can plot these points on the graph.

### 6. **Sketch the parabola:**
   - Draw a smooth curve through these points, ensuring the parabola opens upwards. The curve should pass through the intercepts and have its lowest point at the vertex.

### 7. **Check symmetry:**
   - Parabolas are symmetric about their vertex. Since the vertex is at \( \left(\frac{1}{2}, -\frac{1}{2}\right) \), the graph will be symmetric around the vertical line \( x = \frac{1}{2} \).

### 8. **Label the graph:**
   - Label the vertex, intercepts, and the axis of symmetry \( x = \frac{1}{2} \) on the graph.

### Visual of the Graph:
Imagine a U-shaped curve that dips slightly below the x-axis at \( \left(\frac{1}{2}, -\frac{1}{2}\right) \) and passes through the origin and the point \( (1, 0) \).

Would you like more details or have any questions about these steps?

### Practice Problems:
1. Graph \( y = x^2 - 4x + 4 \) and identify the vertex.
2. Find the vertex and intercepts of \( y = -3x^2 + 6x - 2 \).
3. Determine the axis of symmetry for \( y = x^2 + 2x + 1 \).
4. Graph \( y = -x^2 + 2x + 1 \).
5. Identify the x-intercepts of \( y = x^2 - 5x + 6 \).
6. Calculate the vertex of \( y = 4x^2 + 8x + 3 \).
7. Sketch the graph of \( y = 2x^2 + 3x - 1 \).
8. Find and graph the intercepts of \( y = -2x^2 + 4x - 6 \).

### Tip:
When graphing quadratic functions, always start by finding the vertex. It's the most crucial point for understanding the shape and position of the parabola.

Learn how to graph the quadratic function y = 2x^2 - 2x with a detailed step-by-step guide. This problem covers key algebraic concepts such as graphing quadratic equations, finding vertex, intercepts, and symmetry properties. Suitable for high school students.

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