The equation of normal to the curve `y=x^(3)-x^(2)-1` at the point whose abscissa is -2, is

Options

A. x+16y+206=0
B. x+16y-206=0
C. x+16y+210=0
D. x+16y-210=0

Answer

To find the equation of the normal to the curve \( y = x^3 - x^2 - 1 \) at the point where the abscissa (x-coordinate) is -2, we can follow these steps:

### Step 1: Find the y-coordinate of the point on the curve
We need to substitute \( x = -2 \) into the equation of the curve to find the corresponding \( y \)-coordinate.

\[ y = (-2)^3 - (-2)^2 - 1 \]
\[ y = -8 - 4 - 1 = -13 \]

So, the point on the curve is \( (-2, -13) \).

### Step 2: Find the derivative of the curve
Next, we need to find the derivative \( \frac{dy}{dx} \) of the curve to determine the slope of the tangent line at the point.

The derivative of \( y = x^3 - x^2 - 1 \) is:

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

### Step 3: Evaluate the derivative at \( x = -2 \)
Now, we substitute \( x = -2 \) into the derivative to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=-2} = 3(-2)^2 - 2(-2) \]
\[ = 3(4) + 4 = 12 + 4 = 16 \]

### Step 4: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Thus, the slope \( m \) of the normal line is:

\[ m = -\frac{1}{16} \]

### Step 5: Use the point-slope form to find the equation of the normal line
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), we can substitute \( (x_1, y_1) = (-2, -13) \) and \( m = -\frac{1}{16} \):

\[ y - (-13) = -\frac{1}{16}(x - (-2)) \]
\[ y + 13 = -\frac{1}{16}(x + 2) \]

### Step 6: Rearranging to standard form
Now, we can rearrange this equation to the standard form \( Ax + By + C = 0 \):

\[ y + 13 = -\frac{1}{16}x - \frac{2}{16} \]
\[ y + 13 = -\frac{1}{16}x - \frac{1}{8} \]

Multiplying through by 16 to eliminate the fraction:

\[ 16y + 208 = -x - 2 \]
\[ x + 16y + 210 = 0 \]

Thus, the equation of the normal line is:

\[ \boxed{x + 16y + 210 = 0} \]