Find the equations of the tangent and the normal to the given curve at the indicated point :
`y = x^(3) - 2x + 7 " at " (1, 6)`

Answer

To find the equations of the tangent and the normal to the curve \( y = x^3 - 2x + 7 \) at the point \( (1, 6) \), we will follow these steps:

### Step 1: Differentiate the curve
We start by finding the derivative of the function to determine the slope of the tangent line.

\[ \frac{dy}{dx} = \frac{d}{dx}(x^3 - 2x + 7) \]

Calculating the derivative:

\[ \frac{dy}{dx} = 3x^2 - 2 \]

### Step 2: Evaluate the derivative at the given point
Next, we substitute \( x = 1 \) into the derivative to find the slope of the tangent line at the point \( (1, 6) \).

\[ \frac{dy}{dx} \bigg|_{x=1} = 3(1)^2 - 2 = 3 - 2 = 1 \]

### Step 3: Write the equation of the tangent line
Now that we have the slope of the tangent line \( m = 1 \) and the point \( (1, 6) \), we can use the point-slope form of the equation of a line:

\[ y - y_1 = m(x - x_1) \]

Substituting the values:

\[ y - 6 = 1(x - 1) \]

Simplifying this equation:

\[ y - 6 = x - 1 \implies y = x + 5 \]

Rearranging to standard form:

\[ x - y + 5 = 0 \]

### Step 4: Write the equation of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, if the slope of the tangent is \( 1 \), the slope of the normal \( M \) is:

\[ M = -\frac{1}{1} = -1 \]

Using the point-slope form again for the normal line:

\[ y - y_1 = M(x - x_1) \]

Substituting the values:

\[ y - 6 = -1(x - 1) \]

Simplifying this equation:

\[ y - 6 = -x + 1 \implies y = -x + 7 \]

Rearranging to standard form:

\[ x + y - 7 = 0 \]

### Final Answers
- The equation of the tangent line is \( x - y + 5 = 0 \).
- The equation of the normal line is \( x + y - 7 = 0 \).