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It looks like you have a series of mathematical problems and questions related to quadratic equations, continuity, and specific functions. I'll help you solve each problem step by step.

### Problem 1: Solve the quadratic equation graphically
You mentioned solving the quadratic equations graphically. Let's start with the first equation.

1. **Quadratic Equation 1:** \(x^2 - 2x - 3 = 0\)

   To solve graphically:
   - **Find the roots:** The roots of the quadratic equation are where the parabola intersects the x-axis. The standard form of a quadratic equation is \(ax^2 + bx + c = 0\).
     - Here, \(a = 1\), \(b = -2\), and \(c = -3\).
     - The roots can be found using the quadratic formula:
       \[
       x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
       \]
       Substituting the values:
       \[
       x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}
       \]
       Therefore, the roots are \(x = 3\) and \(x = -1\).

   - **Parabola shape:** Since \(a > 0\), the parabola opens upwards.
   - **Vertex:** The vertex of the parabola can be found using \(x = -\frac{b}{2a}\):
     \[
     x = \frac{2}{2 \times 1} = 1
     \]
     Substituting \(x = 1\) back into the equation to find the y-coordinate:
     \[
     y = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4
     \]
     So, the vertex is at \((1, -4)\).
   - **Graph:** The parabola opens upward with roots at \(x = -1\) and \(x = 3\), and the vertex at \((1, -4)\).

2. **Quadratic Equation 2:** \(x^2 + 7x + 12 = 0\)

   - **Find the roots:** 
     - Here, \(a = 1\), \(b = 7\), and \(c = 12\).
     - Using the quadratic formula:
       \[
       x = \frac{-7 \pm \sqrt{7^2 - 4(1)(12)}}{2(1)} = \frac{-7 \pm \sqrt{49 - 48}}{2} = \frac{-7 \pm \sqrt{1}}{2}
       \]
       Therefore, the roots are \(x = -4\) and \(x = -3\).

   - **Parabola shape:** Since \(a > 0\), the parabola opens upwards.
   - **Vertex:** \(x = -\frac{b}{2a} = -\frac{7}{2} = -3.5\).
     Substituting \(x = -3.5\) back into the equation to find the y-coordinate:
     \[
     y = (-3.5)^2 + 7(-3.5) + 12 = 12.25 - 24.5 + 12 = -0.25
     \]
     So, the vertex is at \((-3.5, -0.25)\).
   - **Graph:** The parabola opens upward with roots at \(x = -4\) and \(x = -3\), and the vertex at \((-3.5, -0.25)\).

3. **Value at Vertex (for a < 0):**
   When \(a < 0\), the parabola opens downward. The value of the quadratic equation at the vertex gives the maximum value.

   - The vertex form of a quadratic equation is \(y = ax^2 + bx + c\).
   - The value at the vertex \(x_v = -\frac{b}{2a}\) would be \(y = f(x_v)\).

### Problem 2: Write the Equation of a Parabola Given the Vertex \((1,8)\) and \(a = 2\)
- The standard form of a parabola with vertex \((h, k)\) is:
  \[
  y = a(x - h)^2 + k
  \]
  Substituting the given values:
  \[
  y = 2(x - 1)^2 + 8
  \]
  This is the equation of the parabola.

### Problem 3: Continuity of a Function at a Point
A function \(f(x)\) is continuous at \(x = 0\) if:
\[
\lim_{{x \to 0}} f(x) = f(0)
\]

### Additional Questions:
1. What is the vertex of the parabola \(y = x^2 - 2x - 3\)?
2. What is the maximum value of \(y = -x^2 + 2x + 3\) and at which point does it occur?
3. Find the axis of symmetry for the parabola \(y = x^2 + 7x + 12\).
4. How do you determine if a function is continuous at a given point?
5. What are the conditions for a quadratic function to have real roots?

**Tip:** When solving quadratic equations graphically, always check the sign of the leading coefficient \(a\) to determine if the parabola opens upwards (minimum) or downwards (maximum).

Explore graphical solutions for quadratic equations, find vertex forms of parabolas, and understand function continuity. This page provides detailed explanations and examples suitable for high school students.

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