Key Concepts and Formulas

• Derivative: The derivative of a function $y(x)$, denoted as $y'(x)$ or $\frac{dy}{dx}$, represents the slope of the tangent line to the curve at the point $x$.
• Touching the x-axis: If a curve touches the x-axis at a point $(x_0, 0)$, then $y(x_0) = 0$ and $y'(x_0) = 0$.
• Local Maxima/Minima: A local maximum or minimum occurs at a critical point $x_0$ where $y'(x_0) = 0$. To determine if it is a maximum, we can check the second derivative: if $y''(x_0) < 0$, it's a local maximum.

Step-by-Step Solution

Step 1: Set up the equations based on the given information.

We are given $y(x) = ax^3 + bx^2 + cx + 5$. The derivative is $y'(x) = 3ax^2 + 2bx + c$.

• Condition 1: The curve touches the x-axis at P(-2, 0). This means $y(-2) = 0$ and $y'(-2) = 0$. $y(-2) = a(-2)^3 + b(-2)^2 + c(-2) + 5 = -8a + 4b - 2c + 5 = 0 \quad (1)$ $y'(-2) = 3a(-2)^2 + 2b(-2) + c = 12a - 4b + c = 0 \quad (2)$
• Condition 2: The curve cuts the y-axis at Q, where y' = 3. The curve cuts the y-axis when $x = 0$. So, $y'(0) = 3$. $y'(0) = 3a(0)^2 + 2b(0) + c = c = 3 \quad (3)$

Step 2: Solve for a, b, and c.

We have a system of three equations with three unknowns: $-8a + 4b - 2c + 5 = 0 \quad (1)$ $12a - 4b + c = 0 \quad (2)$ $c = 3 \quad (3)$

Substitute $c = 3$ into equations (1) and (2): $-8a + 4b - 2(3) + 5 = 0 \Rightarrow -8a + 4b - 1 = 0 \quad (4)$ $12a - 4b + 3 = 0 \quad (5)$

Add equations (4) and (5) to eliminate $b$: $(-8a + 4b - 1) + (12a - 4b + 3) = 0$ $4a + 2 = 0$ $a = -\frac{1}{2}$

Substitute $a = -\frac{1}{2}$ into equation (5): $12\left(-\frac{1}{2}\right) - 4b + 3 = 0$ $-6 - 4b + 3 = 0$ $-3 - 4b = 0$ $b = -\frac{3}{4}$

So, $a = -\frac{1}{2}$, $b = -\frac{3}{4}$, and $c = 3$.

Step 3: Find the expression for y(x) and y'(x).

Now we have $y(x) = -\frac{1}{2}x^3 - \frac{3}{4}x^2 + 3x + 5$ and $y'(x) = -\frac{3}{2}x^2 - \frac{3}{2}x + 3$.

Step 4: Find the critical points.

Set $y'(x) = 0$: $-\frac{3}{2}x^2 - \frac{3}{2}x + 3 = 0$ Multiply by $-\frac{2}{3}$: $x^2 + x - 2 = 0$ $(x + 2)(x - 1) = 0$ The critical points are $x = -2$ and $x = 1$.

Step 5: Find the second derivative y''(x).

$y''(x) = -3x - \frac{3}{2}$

Step 6: Determine if the critical points are local maxima or minima using the second derivative test.

• At $x = -2$: $y''(-2) = -3(-2) - \frac{3}{2} = 6 - \frac{3}{2} = \frac{9}{2} > 0$, so $x = -2$ is a local minimum.
• At $x = 1$: $y''(1) = -3(1) - \frac{3}{2} = -3 - \frac{3}{2} = -\frac{9}{2} < 0$, so $x = 1$ is a local maximum.

Step 7: Calculate the local maximum value.

The local maximum value occurs at $x = 1$: $y(1) = -\frac{1}{2}(1)^3 - \frac{3}{4}(1)^2 + 3(1) + 5 = -\frac{1}{2} - \frac{3}{4} + 3 + 5 = -\frac{2}{4} - \frac{3}{4} + 8 = -\frac{5}{4} + \frac{32}{4} = \frac{27}{4}$

Common Mistakes & Tips

• Remember that "touching" the x-axis implies both $y(x_0) = 0$ and $y'(x_0) = 0$.
• Be careful with signs when solving the system of equations.
• Use the second derivative test to correctly identify local maxima and minima.

Summary

We used the given conditions to set up a system of equations, solved for the coefficients of the cubic polynomial, found the critical points by setting the first derivative to zero, and then used the second derivative test to determine that the local maximum occurs at $x=1$. The local maximum value is then $y(1) = \frac{27}{4}$.

Final Answer

The final answer is \boxed{\frac{27}{4}}, which corresponds to option (A).