Let f left parenthesis x right parenthesis equals 2 x cubed minus 6 x squared minus 90 x minus 3.
a. Find all points on the graph of​ f(x) at which the tangent line is horizontal.
b. Find all points on the graph of​ f(x) at which the tangent line has slope 54.
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Part 1
a. The tangent line is horizontal at the​ point(s)

enter your response here.
​(Type an ordered pair. Use a comma to separate answers as​ needed.)

To find the points on the graph of \( f(x) = 2x^3 - 6x^2 - 90x - 3 \) where the tangent line is horizontal, we need to find where the derivative of the function equals zero.

a. **Finding where the tangent line is horizontal:**

1. Calculate the first derivative of \( f(x) \):
   \[
   f'(x) = \frac{d}{dx}(2x^3 - 6x^2 - 90x - 3)
   \]
   Using the power rule:
   \[
   f'(x) = 6x^2 - 12x - 90
   \]

2. Set the first derivative equal to zero to find the critical points:
   \[
   6x^2 - 12x - 90 = 0
   \]

3. Simplify the equation:
   \[
   x^2 - 2x - 15 = 0
   \]

4. Factor the quadratic equation:
   \[
   (x - 5)(x + 3) = 0
   \]

5. Solve for \( x \):
   \[
   x = 5 \quad \text{or} \quad x = -3
   \]

6. Find the corresponding \( y \)-values by substituting \( x \) back into the original function \( f(x) \):
   \[
   f(5) = 2(5)^3 - 6(5)^2 - 90(5) - 3 = 2(125) - 6(25) - 450 - 3 = 250 - 150 - 450 - 3 = -353
   \]
   \[
   f(-3) = 2(-3)^3 - 6(-3)^2 - 90(-3) - 3 = 2(-27) - 6(9) + 270 - 3 = -54 - 54 + 270 - 3 = 159
   \]

Thus, the points on the graph where the tangent line is horizontal are:
\[
(5, -353) \quad \text{and} \quad (-3, 159)
\]

b. **Finding where the tangent line has a slope of 54:**

1. Set the first derivative equal to 54:
   \[
   6x^2 - 12x - 90 = 54
   \]

2. Simplify the equation:
   \[
   6x^2 - 12x - 144 = 0
   \]
   \[
   x^2 - 2x - 24 = 0
   \]

3. Factor the quadratic equation:
   \[
   (x - 6)(x + 4) = 0
   \]

4. Solve for \( x \):
   \[
   x = 6 \quad \text{or} \quad x = -4
   \]

5. Find the corresponding \( y \)-values by substituting \( x \) back into the original function \( f(x) \):
   \[
   f(6) = 2(6)^3 - 6(6)^2 - 90(6) - 3 = 2(216) - 6(36) - 540 - 3 = 432 - 216 - 540 - 3 = -327
   \]
   \[
   f(-4) = 2(-4)^3 - 6(-4)^2 - 90(-4) - 3 = 2(-64) - 6(16) + 360 - 3 = -128 - 96 + 360 - 3 = 133
   \]

Thus, the points on the graph where the tangent line has a slope of 54 are:
\[
(6, -327) \quad \text{and} \quad (-4, 133)
\]

**Summary:**
- The points where the tangent line is horizontal are \((5, -353)\) and \((-3, 159)\).
- The points where the tangent line has a slope of 54 are \((6, -327)\) and \((-4, 133)\).

Would you like more details or have any questions? Here are some related questions for further exploration:

1. How do you find the second derivative of \( f(x) \) and what does it tell you about the concavity of the function?
2. What are the critical points of the function \( f(x) \)?
3. How do you determine if the critical points are local minima, maxima, or saddle points?
4. What is the inflection point of \( f(x) \)?
5. How can you graph the function \( f(x) \) to visualize these points?
6. What is the significance of the discriminant in the quadratic equations solved above?
7. How do changes in the coefficients of \( f(x) \) affect the critical points and points of horizontal tangency?
8. How would you solve the same problem using a graphing calculator or computer software?

**Tip:** When solving for points where the tangent line has a specific slope, always ensure to substitute the critical \( x \)-values back into the original function to find the corresponding \( y \)-values accurately.

Learn how to find points on the graph of f(x) = 2x^3 - 6x^2 - 90x - 3 where the tangent line is horizontal or has a slope of 54. This problem involves calculus concepts such as derivatives, critical points, and the power rule. Suitable for advanced high school students studying calculus.

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