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).