Show that there lies a point on the curve `f(x)=x^(2)-4x +3` between (1,0) and (3,0) where tangent drawn is parallel to x-axis. Find its coordinates.

Answer

To solve the problem, we need to show that there exists a point on the curve \( f(x) = x^2 - 4x + 3 \) between the points \( (1,0) \) and \( (3,0) \) where the tangent is parallel to the x-axis. A tangent parallel to the x-axis indicates that the derivative of the function at that point is zero.

### Step-by-Step Solution:

1. **Find the derivative of the function**:
\[ f(x) = x^2 - 4x + 3 \]
To find the derivative, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^2 - 4x + 3) = 2x - 4 \]

2. **Set the derivative equal to zero**:
Since we want the tangent to be parallel to the x-axis, we set the derivative equal to zero:
\[ f'(x) = 0 \implies 2x - 4 = 0 \]

3. **Solve for \( x \)**:
Solving the equation \( 2x - 4 = 0 \):
\[ 2x = 4 \implies x = 2 \]

4. **Find the corresponding \( y \)-coordinate**:
Now we need to find the \( y \)-coordinate of the point on the curve when \( x = 2 \):
\[ f(2) = 2^2 - 4 \cdot 2 + 3 \]
Calculating this:
\[ f(2) = 4 - 8 + 3 = -1 \]

5. **State the coordinates of the point**:
Therefore, the coordinates of the point where the tangent is parallel to the x-axis are:
\[ (2, -1) \]

6. **Verify that the point lies between (1,0) and (3,0)**:
The point \( (2, -1) \) lies between \( (1,0) \) and \( (3,0) \) on the x-axis.

### Final Answer:
The coordinates of the point on the curve where the tangent is parallel to the x-axis are \( (2, -1) \).

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