At which point the slope to tangent is zero for the curvey `y=x^2-6x+8` ?

Options

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

Answer

To find the point at which the slope of the tangent to the curve \( y = x^2 - 6x + 8 \) is zero, we can follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y \) with respect to \( x \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = \frac{d}{dx}(x^2 - 6x + 8) \]

### Step 2: Apply the power rule
Using the power rule of differentiation, we differentiate each term:

- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( -6x \) is \( -6 \).
- The derivative of the constant \( 8 \) is \( 0 \).

Thus, we have:

\[ \frac{dy}{dx} = 2x - 6 \]

### Step 3: Set the derivative equal to zero
To find the points where the slope of the tangent is zero, we set the derivative equal to zero:

\[ 2x - 6 = 0 \]

### Step 4: Solve for \( x \)
Now, we solve for \( x \):

\[ 2x = 6 \implies x = 3 \]

### Step 5: Find the corresponding \( y \) value
Next, we substitute \( x = 3 \) back into the original equation to find the corresponding \( y \) value:

\[ y = (3)^2 - 6(3) + 8 \]

Calculating this gives:

\[ y = 9 - 18 + 8 = -1 \]

### Step 6: State the point
Thus, the point at which the slope of the tangent is zero is:

\[ (3, -1) \]

### Final Answer
The slope of the tangent is zero at the point \( (3, -1) \).

---