Find the coordinates of the vertex of the parabola `y=x^2-4x+1` by making use of the fact that at the vertex, the slope of the tangent is zero.

Options

A. (2,-3)
B. (3,-2)
C. (-1,-3)
D. (-2,-3)

Answer

To find the coordinates of the vertex of the parabola given by the equation \( y = x^2 - 4x + 1 \), we will use the fact that at the vertex, the slope of the tangent (the derivative) is zero. Here’s a step-by-step solution:

### Step 1: Find the derivative of the function
We start by differentiating the function \( y = x^2 - 4x + 1 \) with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(1) \]

Calculating the derivatives, we get:

\[ \frac{dy}{dx} = 2x - 4 + 0 = 2x - 4 \]

### Step 2: Set the derivative equal to zero
To find the x-coordinate of the vertex, we set the derivative equal to zero:

\[ 2x - 4 = 0 \]

### Step 3: Solve for \( x \)
Now, we solve for \( x \):

\[ 2x = 4 \implies x = 2 \]

### Step 4: Find the corresponding \( y \)-coordinate
Next, we substitute \( x = 2 \) back into the original equation to find the corresponding \( y \)-coordinate:

\[ y = (2)^2 - 4(2) + 1 \]

Calculating this gives:

\[ y = 4 - 8 + 1 = -3 \]

### Step 5: Write the coordinates of the vertex
Thus, the coordinates of the vertex of the parabola are:

\[ (2, -3) \]

### Summary
The vertex of the parabola \( y = x^2 - 4x + 1 \) is at the point \( (2, -3) \).

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