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The problem asks for two curve tracing processes based on the quadratic functions provided. Let's work through both of them step by step:

### 1. \( y = x^2 + 6x - 9 \)

#### Step 1: Find the key points
- **Vertex**: The vertex of a quadratic function \( y = ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \).
  - Here, \( a = 1 \), \( b = 6 \), and \( c = -9 \).
  - \( x = -\frac{6}{2(1)} = -3 \).
  - To find the y-coordinate of the vertex, substitute \( x = -3 \) back into the equation:
    - \( y = (-3)^2 + 6(-3) - 9 = 9 - 18 - 9 = -18 \).
  - **Vertex**: \( (-3, -18) \).

- **Axis of Symmetry**: The axis of symmetry is \( x = -3 \).

- **Y-Intercept**: When \( x = 0 \), \( y = (0)^2 + 6(0) - 9 = -9 \).
  - **Y-intercept**: \( (0, -9) \).

- **X-Intercepts**: To find the x-intercepts, solve \( x^2 + 6x - 9 = 0 \) using the quadratic formula:
  - \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4(1)(-9)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 36}}{2} = \frac{-6 \pm \sqrt{72}}{2} = \frac{-6 \pm 6\sqrt{2}}{2} \).
  - **Roots**: \( x = -3 \pm 3\sqrt{2} \), approximately \( x \approx -7.24 \) and \( x \approx 1.24 \).

#### Step 2: Plot key features
- The parabola opens upwards (since \( a = 1 > 0 \)).
- The vertex is \( (-3, -18) \), the axis of symmetry is \( x = -3 \), and the y-intercept is \( (0, -9) \).
- The x-intercepts are \( x \approx -7.24 \) and \( x \approx 1.24 \).

#### Step 3: Sketch the curve
- Draw the curve with the vertex at \( (-3, -18) \), crossing the y-axis at \( (0, -9) \), and passing through the x-intercepts.

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### 2. \( y = 2x^2 - 7x + 8 \)

#### Step 1: Find the key points
- **Vertex**: Using the same formula, \( x = -\frac{b}{2a} \).
  - Here, \( a = 2 \), \( b = -7 \), and \( c = 8 \).
  - \( x = -\frac{-7}{2(2)} = \frac{7}{4} = 1.75 \).
  - Substitute \( x = 1.75 \) into the equation to find the y-coordinate:
    - \( y = 2(1.75)^2 - 7(1.75) + 8 = 2(3.0625) - 12.25 + 8 = 6.125 - 12.25 + 8 = 1.875 \).
  - **Vertex**: \( (1.75, 1.875) \).

- **Axis of Symmetry**: The axis of symmetry is \( x = 1.75 \).

- **Y-Intercept**: When \( x = 0 \), \( y = 2(0)^2 - 7(0) + 8 = 8 \).
  - **Y-intercept**: \( (0, 8) \).

- **X-Intercepts**: To find the x-intercepts, solve \( 2x^2 - 7x + 8 = 0 \) using the quadratic formula:
  - \( x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(8)}}{2(2)} = \frac{7 \pm \sqrt{49 - 64}}{4} = \frac{7 \pm \sqrt{-15}}{4} \).
  - The discriminant is negative, so there are **no real x-intercepts** (the curve does not cross the x-axis).

#### Step 2: Plot key features
- The parabola opens upwards (since \( a = 2 > 0 \)).
- The vertex is \( (1.75, 1.875) \), the axis of symmetry is \( x = 1.75 \), and the y-intercept is \( (0, 8) \).
- Since there are no real x-intercepts, the parabola does not cross the x-axis.

#### Step 3: Sketch the curve
- Draw the curve with the vertex at \( (1.75, 1.875) \), passing through the y-axis at \( (0, 8) \).

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Would you like further details or have any questions?

### Related Questions:
1. How does the discriminant determine the number of x-intercepts in a quadratic equation?
2. How do you convert a quadratic function into vertex form?
3. What role does the coefficient of \( x^2 \) play in the shape of a parabola?
4. Can a quadratic function have complex roots and still have a real graph?
5. How do changes to the constants in a quadratic equation affect its graph?

### Tip:
When graphing parabolas, always start with the vertex, as it gives the critical point from which the rest of the curve is symmetrical.

Learn how to solve and graph the quadratic equations y = x^2 + 6x - 9 and y = 2x^2 - 7x + 8. This solution includes finding the vertex, axis of symmetry, y-intercept, and x-intercepts, as well as graphing key features of the quadratic functions. Suitable for students in Grades 9-11.

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