Find the equation of parabola whose axis is parallel to x-axis and which passes through the points
(-2, I) (1,2) (-1. 3)

Answer

To find the equation of the parabola whose axis is parallel to the x-axis and which passes through the points (-2, 1), (1, 2), and (-1, 3), we can follow these steps:

### Step 1: Write the general equation of the parabola
The equation of a parabola with its axis parallel to the x-axis can be expressed as:
\[ y = ax^2 + bx + c \]
Let this be Equation (1).

### Step 2: Substitute the first point (-2, 1) into the equation
Substituting \( x = -2 \) and \( y = 1 \) into Equation (1):
\[ 1 = a(-2)^2 + b(-2) + c \]
This simplifies to:
\[ 1 = 4a - 2b + c \]
Rearranging gives us:
\[ 4a - 2b + c = 1 \quad \text{(Equation 2)} \]

### Step 3: Substitute the second point (1, 2) into the equation
Substituting \( x = 1 \) and \( y = 2 \) into Equation (1):
\[ 2 = a(1)^2 + b(1) + c \]
This simplifies to:
\[ 2 = a + b + c \]
Rearranging gives us:
\[ a + b + c = 2 \quad \text{(Equation 3)} \]

### Step 4: Substitute the third point (-1, 3) into the equation
Substituting \( x = -1 \) and \( y = 3 \) into Equation (1):
\[ 3 = a(-1)^2 + b(-1) + c \]
This simplifies to:
\[ 3 = a - b + c \]
Rearranging gives us:
\[ a - b + c = 3 \quad \text{(Equation 4)} \]

### Step 5: Solve the system of equations
Now we have a system of three equations:
1. \( 4a - 2b + c = 1 \) (Equation 2)
2. \( a + b + c = 2 \) (Equation 3)
3. \( a - b + c = 3 \) (Equation 4)

We can solve these equations step by step.

**From Equation (3)**:
\[ c = 2 - a - b \quad \text{(Substituting for c)} \]

**Substituting c into Equation (4)**:
\[ a - b + (2 - a - b) = 3 \]
This simplifies to:
\[ 2 - 2b = 3 \implies -2b = 1 \implies b = -\frac{1}{2} \]

**Substituting b into Equation (3)**:
\[ a - \frac{1}{2} + c = 2 \implies c = 2 - a + \frac{1}{2} = \frac{5}{2} - a \]

**Substituting b and c into Equation (2)**:
\[ 4a - 2(-\frac{1}{2}) + \left(\frac{5}{2} - a\right) = 1 \]
This simplifies to:
\[ 4a + 1 + \frac{5}{2} - a = 1 \]
Combining like terms gives:
\[ 3a + \frac{7}{2} = 1 \implies 3a = 1 - \frac{7}{2} = -\frac{5}{2} \implies a = -\frac{5}{6} \]

**Now substituting a back to find c**:
\[ c = \frac{5}{2} - \left(-\frac{5}{6}\right) = \frac{5}{2} + \frac{5}{6} = \frac{15}{6} + \frac{5}{6} = \frac{20}{6} = \frac{10}{3} \]

### Step 6: Write the final equation of the parabola
Now we have:
- \( a = -\frac{5}{6} \)
- \( b = -\frac{1}{2} \)
- \( c = \frac{10}{3} \)

Substituting these values into Equation (1):
\[ y = -\frac{5}{6}x^2 - \frac{1}{2}x + \frac{10}{3} \]

### Final Answer
The equation of the parabola is:
\[ y = -\frac{5}{6}x^2 - \frac{1}{2}x + \frac{10}{3} \]