Find the co-ordinates of the point at which the tangent to the curve given by `f(x)=x^(2)-6x+1` is parallel to the chord joining the points (1, -4) and (3, -8).

Answer

To find the coordinates of the point at which the tangent to the curve \( f(x) = x^2 - 6x + 1 \) is parallel to the chord joining the points (1, -4) and (3, -8), we can follow these steps:

### Step 1: Find the slope of the chord
The slope of the chord joining the points \( (1, -4) \) and \( (3, -8) \) can be calculated using the formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates:
\[ \text{slope} = \frac{-8 - (-4)}{3 - 1} = \frac{-8 + 4}{2} = \frac{-4}{2} = -2 \]

### Step 2: Find the derivative of the function
Next, we find the derivative of the function \( f(x) = x^2 - 6x + 1 \) to determine the slope of the tangent line:
\[ f'(x) = \frac{d}{dx}(x^2 - 6x + 1) = 2x - 6 \]

### Step 3: Set the derivative equal to the slope of the chord
Since the tangent is parallel to the chord, we set the derivative equal to the slope of the chord:
\[ 2x - 6 = -2 \]

### Step 4: Solve for \( x \)
Now, we solve for \( x \):
\[ 2x = -2 + 6 \]
\[ 2x = 4 \]
\[ x = 2 \]

### Step 5: Find the corresponding \( y \) coordinate
Now we substitute \( x = 2 \) back into the original function to find the corresponding \( y \) coordinate:
\[ f(2) = 2^2 - 6(2) + 1 = 4 - 12 + 1 = -7 \]

### Conclusion
Thus, the coordinates of the point at which the tangent to the curve is parallel to the chord are:
\[ \boxed{(2, -7)} \]