Find all the values of x where the tangent line is horizontal.
f left parenthesis x right parenthesis equals 2 x cubed plus 39 x squared plus 252 x plus 9


To find where the tangent line is horizontal, we need to find the values of \( x \) where the derivative of the function is equal to zero. The derivative represents the slope of the tangent line, and a horizontal tangent line occurs when the slope is zero.

The given function is:
\[
f(x) = 2x^3 + 39x^2 + 252x + 9
\]

### Step 1: Find the derivative of \( f(x) \)

We differentiate \( f(x) \) with respect to \( x \):

\[
f'(x) = \frac{d}{dx}(2x^3 + 39x^2 + 252x + 9)
\]
\[
f'(x) = 6x^2 + 78x + 252
\]

### Step 2: Set the derivative equal to zero

Now, set \( f'(x) = 0 \) to find the points where the tangent line is horizontal:

\[
6x^2 + 78x + 252 = 0
\]

### Step 3: Solve the quadratic equation

We can simplify the equation by dividing everything by 6:

\[
x^2 + 13x + 42 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \( x^2 + 13x + 42 = 0 \), we have:
- \( a = 1 \)
- \( b = 13 \)
- \( c = 42 \)

Substitute these values into the quadratic formula:

\[
x = \frac{-13 \pm \sqrt{13^2 - 4(1)(42)}}{2(1)}
\]
\[
x = \frac{-13 \pm \sqrt{169 - 168}}{2}
\]
\[
x = \frac{-13 \pm \sqrt{1}}{2}
\]
\[
x = \frac{-13 \pm 1}{2}
\]

This gives two solutions:
\[
x = \frac{-13 + 1}{2} = -6
\]
\[
x = \frac{-13 - 1}{2} = -7
\]

### Conclusion

The values of \( x \) where the tangent line is horizontal are:
\[
x = -6 \quad \text{and} \quad x = -7
\]

Would you like any details on this process, or do you have further questions?

Here are 5 related questions for further exploration:
1. What is the geometric interpretation of a horizontal tangent line?
2. How does the second derivative of the function help determine the concavity at these points?
3. Can the same approach be used to find vertical tangents?
4. What happens if we compute the tangent at values of \( x \) not where \( f'(x) = 0 \)?
5. How would the function behave around the points \( x = -6 \) and \( x = -7 \)?

**Tip**: Always check if simplifying the quadratic equation (like dividing all terms by a constant) can make solving easier.

Find all the values of x where the tangent line is horizontal. f(x) = 2x^3 + 39x^2 + 252x + 9

Learn how to find the values of x where the tangent line is horizontal for the cubic function f(x) = 2x^3 + 39x^2 + 252x + 9. This solution includes finding the derivative, solving a quadratic equation, and understanding the conditions for a horizontal tangent. Suitable for students in Grades 10-12.

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