Find a point on the curve `y=(x-3)^2` , where the tangent is parallel to the line joining `(4,\ 1)` and `(3,\ 0)` .

Explanation

To find the point on the curve \( y = (x - 3)^2 \) where the tangent is parallel to the line joining the points \( (4, 1) \) and \( (3, 0) \), we can follow these steps:

### Step 1: Find the slope of the line joining the points \( (4, 1) \) and \( (3, 0) \).

The slope \( m \) of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substituting the given points \( (4, 1) \) and \( (3, 0) \):
\[ m = \frac{0 - 1}{3 - 4} = \frac{-1}{-1} = 1 \]

### Step 2: Differentiate the curve \( y = (x - 3)^2 \) to find the slope of the tangent.

To find the slope of the tangent line to the curve at any point, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2(x - 3) \]

### Step 3: Set the derivative equal to the slope of the line.

Since we want the tangent to be parallel to the line, we set the derivative equal to the slope we found in Step 1:
\[ 2(x - 3) = 1 \]

### Step 4: Solve for \( x \).

Now, we solve the equation:
\[ 2(x - 3) = 1 \]
\[ x - 3 = \frac{1}{2} \]
\[ x = 3 + \frac{1}{2} = \frac{7}{2} \]

### Step 5: Find the corresponding \( y \) value on the curve.

Now that we have \( x = \frac{7}{2} \), we can find the corresponding \( y \) value by substituting \( x \) back into the equation of the curve:
\[ y = \left(\frac{7}{2} - 3\right)^2 = \left(\frac{7}{2} - \frac{6}{2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 6: State the point.

Thus, the point on the curve where the tangent is parallel to the line joining \( (4, 1) \) and \( (3, 0) \) is:
\[ \left(\frac{7}{2}, \frac{1}{4}\right) \]