The problem asks for the values of $x$ at which the tangent line to the graph of $f(x) = 2x^3 + 3x^2 - 72x$ is parallel to the line $y = -1.7$.

Key points:

• The slope of the line $y = -1.7$ is constant and equal to 0, since it is a horizontal line.
• To find where the tangent line to the function $f(x)$ is parallel to this line, we need to determine where the derivative $f'(x)$ equals 0 (because parallel lines must have the same slope).

Steps:

1. Find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 72x)$ $f'(x) = 6x^2 + 6x - 72$

2. Set the derivative equal to the slope of the line $y = -1.7$: Since the slope of the line is 0, we set: $6x^2 + 6x - 72 = 0$

3. Solve the quadratic equation: $x^2 + x - 12 = 0 \quad \text{(dividing the equation by 6)}$ Factor the quadratic: $(x + 4)(x - 3) = 0$ Therefore, $x = -4$ and $x = 3$.

Conclusion:

The tangent line to the graph of $f(x) = 2x^3 + 3x^2 - 72x$ is parallel to the line $y = -1.7$ at $x = -4$ and $x = 3$.

Would you like more details on any part of the solution?

Here are five related questions to expand on this:

1. What is the geometric meaning of a tangent line being parallel to another line?
2. How do you differentiate polynomials using the power rule?
3. Why does the quadratic formula solve any quadratic equation?
4. Can there be more than two solutions to a cubic function's tangent line being parallel to a given line?
5. How do the roots of the derivative relate to the critical points of a function?

Tip: When solving a quadratic equation, always check if factoring is possible before using the quadratic formula. It saves time!