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.