Use 1 whole sheet of white pad paper and a graphing paper to find the solution set and graph of y > 2x2 - 3x+ 1

To solve and graph the inequality \( y > 2x^2 - 3x + 1 \), follow these steps:

### 1. Solve the inequality algebraically:
The given inequality is:
\[
y > 2x^2 - 3x + 1
\]
- This is a quadratic inequality, where the right-hand side is a parabola.
- The first step is to treat the expression as an equation \( y = 2x^2 - 3x + 1 \) and find the critical points (vertex and x-intercepts).

### 2. Find the vertex:
For a quadratic equation of the form \( y = ax^2 + bx + c \), the x-coordinate of the vertex is found using:
\[
x_{\text{vertex}} = \frac{-b}{2a}
\]
Here, \( a = 2 \), \( b = -3 \), and \( c = 1 \).

\[
x_{\text{vertex}} = \frac{-(-3)}{2(2)} = \frac{3}{4}
\]
Now, substitute \( x_{\text{vertex}} = \frac{3}{4} \) into the equation to find the y-coordinate of the vertex:
\[
y_{\text{vertex}} = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 1
\]
Calculating the result:
\[
y_{\text{vertex}} = 2\left(\frac{9}{16}\right) - 3\left(\frac{3}{4}\right) + 1 = \frac{18}{16} - \frac{9}{4} + 1 = \frac{9}{8} - \frac{9}{4} + 1
\]
\[
y_{\text{vertex}} = \frac{9}{8} - \frac{18}{8} + \frac{8}{8} = \frac{-1}{8}
\]
Thus, the vertex is at \( \left( \frac{3}{4}, -\frac{1}{8} \right) \).

### 3. Plot the parabola:
- The parabola opens upwards because \( a = 2 \) (positive coefficient).
- The vertex \( \left( \frac{3}{4}, -\frac{1}{8} \right) \) is the minimum point.

### 4. Find the x-intercepts (if needed for graphing):
The x-intercepts occur when \( y = 0 \), so solve:
\[
0 = 2x^2 - 3x + 1
\]
Using the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}
\]
So the x-intercepts are:
\[
x_1 = \frac{3 + 1}{4} = 1, \quad x_2 = \frac{3 - 1}{4} = \frac{1}{2}
\]
Thus, the x-intercepts are \( x = 1 \) and \( x = \frac{1}{2} \).

### 5. Graph the inequality:
- Plot the parabola using the vertex \( \left( \frac{3}{4}, -\frac{1}{8} \right) \) and the x-intercepts \( x = 1 \) and \( x = \frac{1}{2} \).
- Since the inequality is \( y > 2x^2 - 3x + 1 \), you need to shade the region **above** the parabola. 
- The boundary parabola \( y = 2x^2 - 3x + 1 \) is not included in the solution set, so draw a dashed curve for the parabola.

### Graphing Tips:
- On the graph paper, plot the points for the vertex, the x-intercepts, and other points as needed to capture the shape of the parabola.
- Use a dashed line to represent the boundary of the inequality, as the points on the parabola do not satisfy the strict inequality \( y > 2x^2 - 3x + 1 \).
- Shade the region above the parabola to represent the solution set.

Let me know if you would like further clarifications or if you want to see a step-by-step graph.

---

Here are 5 related questions to expand your understanding:
1. How does changing the coefficient of \(x^2\) affect the direction and width of a parabola?
2. What is the significance of the vertex in a quadratic inequality?
3. How would the graph change if the inequality was \(y \geq 2x^2 - 3x + 1\)?
4. What are the differences between graphing a strict inequality and a non-strict inequality?
5. How do you determine the region to shade when solving inequalities graphically?

**Tip**: Always pay attention to whether the inequality is strict or non-strict to know whether the boundary line or curve should be dashed or solid.

Use 1 whole sheet of white pad paper and a graphing paper to find the solution set and graph of y > 2x^2 - 3x + 1

Explore how to solve and graph the quadratic inequality y > 2x^2 - 3x + 1 with a step-by-step guide. This problem covers key algebraic concepts like quadratic inequalities, vertex calculation, and graphing a parabola. Learn the methods suitable for students in Grades 9-11.

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