Solve graphically and compare your answer with algebraic solution either by factorization or formula method:
(i) `y=x^(2)-5x+6`
(ii) `y=-x^(2)+2x+3`
(iii) `y=x^(2)-4x+4`
(iv) `y=x^(2)-x-6`
(v) `y=x^(2)-6x+9`
(vi) `y=-x^(2)-x+12`
(vii) `y=x^(2)-4x+5=0`
(viii) `y=x^(2)+2x+2=0`.

Answer

To solve the quadratic equations graphically and compare the results with algebraic solutions, we will follow these steps for each equation:

### Step-by-Step Solution

1. **Equation (i): \(y = x^2 - 5x + 6\)**

**Algebraic Solution:**
- Factor the equation:
\[ y = x^2 - 5x + 6 = (x - 2)(x - 3) \]
- Set \(y = 0\):
\[ (x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3 \]
- Roots: \(x = 2\) and \(x = 3\).

**Graphical Solution:**
- Plot the parabola based on the factored form.
- The parabola intersects the x-axis at \(x = 2\) and \(x = 3\).

2. **Equation (ii): \(y = -x^2 + 2x + 3\)**

**Algebraic Solution:**
- Factor the equation:
\[ y = -x^2 + 2x + 3 = -(x^2 - 2x - 3) = -(x - 3)(x + 1) \]
- Set \(y = 0\):
\[ -(x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1 \]
- Roots: \(x = 3\) and \(x = -1\).

**Graphical Solution:**
- The parabola opens downwards and intersects the x-axis at \(x = 3\) and \(x = -1\).

3. **Equation (iii): \(y = x^2 - 4x + 4\)**

**Algebraic Solution:**
- Factor the equation:
\[ y = (x - 2)^2 \]
- Set \(y = 0\):
\[ (x - 2)^2 = 0 \implies x = 2 \]
- Root: \(x = 2\) (double root).

**Graphical Solution:**
- The parabola touches the x-axis at \(x = 2\).

4. **Equation (iv): \(y = x^2 - x - 6\)**

**Algebraic Solution:**
- Factor the equation:
\[ y = (x - 3)(x + 2) \]
- Set \(y = 0\):
\[ (x - 3)(x + 2) = 0 \implies x = 3 \text{ or } x = -2 \]
- Roots: \(x = 3\) and \(x = -2\).

**Graphical Solution:**
- The parabola intersects the x-axis at \(x = 3\) and \(x = -2\).

5. **Equation (v): \(y = x^2 - 6x + 9\)**

**Algebraic Solution:**
- Factor the equation:
\[ y = (x - 3)^2 \]
- Set \(y = 0\):
\[ (x - 3)^2 = 0 \implies x = 3 \]
- Root: \(x = 3\) (double root).

**Graphical Solution:**
- The parabola touches the x-axis at \(x = 3\).

6. **Equation (vi): \(y = -x^2 - x + 12\)**

**Algebraic Solution:**
- Factor the equation:
\[ y = -1(x^2 + x - 12) = -1(x - 3)(x + 4) \]
- Set \(y = 0\):
\[ -(x - 3)(x + 4) = 0 \implies x = 3 \text{ or } x = -4 \]
- Roots: \(x = 3\) and \(x = -4\).

**Graphical Solution:**
- The parabola opens downwards and intersects the x-axis at \(x = 3\) and \(x = -4\).

7. **Equation (vii): \(y = x^2 - 4x + 5\)**

**Algebraic Solution:**
- Use the quadratic formula:
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} \]
- Roots are imaginary.

**Graphical Solution:**
- The parabola does not intersect the x-axis.

8. **Equation (viii): \(y = x^2 + 2x + 2\)**

**Algebraic Solution:**
- Use the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} \]
- Roots are imaginary.

**Graphical Solution:**
- The parabola does not intersect the x-axis.