The tangent to `y=ax^(2)+bx+c` at `(1,2)` is parallel to the normal at the point `(-2,2)` on the same curve. Find the value of `3a-b+c`

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript while ensuring clarity in each step.

### Step 1: Establish the equation of the curve
The curve is given by the equation:
\[ y = ax^2 + bx + c \]

### Step 2: Use the point (1, 2) on the curve
Since the point (1, 2) lies on the curve, we can substitute \( x = 1 \) and \( y = 2 \) into the equation:
\[ 2 = a(1)^2 + b(1) + c \]
This simplifies to:
\[ 2 = a + b + c \]
Let's label this as Equation (1):
\[ a + b + c = 2 \]

### Step 3: Use the point (-2, 2) on the curve
Similarly, since the point (-2, 2) also lies on the curve, we substitute \( x = -2 \) and \( y = 2 \):
\[ 2 = a(-2)^2 + b(-2) + c \]
This simplifies to:
\[ 2 = 4a - 2b + c \]
Let's label this as Equation (2):
\[ 4a - 2b + c = 2 \]

### Step 4: Set up the equations for comparison
Now we have two equations:
1. \( a + b + c = 2 \) (Equation 1)
2. \( 4a - 2b + c = 2 \) (Equation 2)

### Step 5: Eliminate \( c \)
To eliminate \( c \), we can subtract Equation (1) from Equation (2):
\[ (4a - 2b + c) - (a + b + c) = 2 - 2 \]
This simplifies to:
\[ 4a - 2b + c - a - b - c = 0 \]
\[ 3a - 3b = 0 \]
From this, we can conclude:
\[ 3a = 3b \quad \Rightarrow \quad a = b \]

### Step 6: Substitute \( b \) with \( a \)
Now that we know \( a = b \), we can substitute \( b \) in Equation (1):
\[ a + a + c = 2 \]
This simplifies to:
\[ 2a + c = 2 \quad \Rightarrow \quad c = 2 - 2a \]

### Step 7: Find the value of \( 3a - b + c \)
Now, we need to find \( 3a - b + c \):
Substituting \( b = a \) and \( c = 2 - 2a \):
\[ 3a - a + (2 - 2a) = 3a - a + 2 - 2a \]
This simplifies to:
\[ (3a - a - 2a) + 2 = 2 \]
Thus:
\[ 3a - b + c = 2 \]

### Final Answer
The value of \( 3a - b + c \) is:
\[ \boxed{2} \]