Find the equation of the normal line to the curve `f(x)=5x^(3)-2x^(2)-3x-1" at " x=-8`

Answer

To find the equation of the normal line to the curve \( f(x) = 5x^3 - 2x^2 - 3x - 1 \) at the point where \( x = -8 \), we will follow these steps:

### Step 1: Find the point on the curve at \( x = -8 \)

We need to calculate \( f(-8) \):

\[ f(-8) = 5(-8)^3 - 2(-8)^2 - 3(-8) - 1 \]

Calculating each term:

- \( 5(-8)^3 = 5 \times (-512) = -2560 \)
- \( -2(-8)^2 = -2 \times 64 = -128 \)
- \( -3(-8) = 24 \)
- \( -1 = -1 \)

Now, summing these values:

\[ f(-8) = -2560 - 128 + 24 - 1 = -2665 \]

Thus, the point on the curve is \( (-8, -2665) \).

### Step 2: Find the derivative \( f'(x) \) to get the slope of the tangent line

The derivative of \( f(x) \) is:

\[ f'(x) = \frac{d}{dx}(5x^3 - 2x^2 - 3x - 1) = 15x^2 - 4x - 3 \]

### Step 3: Calculate the slope of the tangent line at \( x = -8 \)

Now we substitute \( x = -8 \) into \( f'(x) \):

\[ f'(-8) = 15(-8)^2 - 4(-8) - 3 \]

Calculating each term:

- \( 15(-8)^2 = 15 \times 64 = 960 \)
- \( -4(-8) = 32 \)
- \( -3 = -3 \)

Now, summing these values:

\[ f'(-8) = 960 + 32 - 3 = 989 \]

So, the slope of the tangent line at \( x = -8 \) is \( 989 \).

### 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:

\[ \text{slope of normal} = -\frac{1}{989} \]

### Step 5: Use the point-slope form to find the equation of the normal line

Using the point-slope form \( y - y_1 = m(x - x_1) \):

\[ y - (-2665) = -\frac{1}{989}(x - (-8)) \]

This simplifies to:

\[ y + 2665 = -\frac{1}{989}(x + 8) \]

### Step 6: Rearranging to standard form

Multiplying through by \( 989 \) to eliminate the fraction:

\[ 989(y + 2665) = -1(x + 8) \]

Expanding both sides:

\[ 989y + 2636935 = -x - 8 \]

Rearranging gives:

\[ x + 989y + 2636943 = 0 \]

### Final Equation

Thus, the equation of the normal line is:

\[ x + 989y + 2636943 = 0 \]

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