Key Concepts and Formulas

• Vertex of a Quadratic: For a quadratic function $f(x) = Ax^2 + Bx + C$, the x-coordinate of the vertex is given by $x_v = -\frac{B}{2A}$.
• Monotonicity and Leading Coefficient: If $A > 0$, the parabola opens upwards, and the function has a local minimum. If $A < 0$, the parabola opens downwards, and the function has a local maximum.
• Relationship between Monotonicity and Vertex: The vertex of a parabola is the point where the function changes from increasing to decreasing (local maximum) or decreasing to increasing (local minimum).

Step-by-Step Solution

Step 1: Analyze f(x) and determine the value of 'a'

We are given $f(x) = ax^2 + 6x - 15$. The problem states that $f(x)$ is increasing on $(-\infty, \frac{3}{4})$ and decreasing on $(\frac{3}{4}, \infty)$. This means that $f(x)$ has a local maximum at $x = \frac{3}{4}$. Since $f(x)$ has a local maximum, the coefficient 'a' must be negative (parabola opens downwards). Also, the x-coordinate of the vertex is $\frac{3}{4}$.

Using the vertex formula $x_v = -\frac{B}{2A}$ for $f(x)$, where $A = a$ and $B = 6$, we have:

$x_v = -\frac{6}{2a}$

We are given that $x_v = \frac{3}{4}$, so:

$\frac{3}{4} = -\frac{6}{2a}$

Step 2: Solve for 'a'

We solve the equation for 'a':

$\frac{3}{4} = -\frac{3}{a}$

Cross-multiplying gives:

$3a = -12$

Dividing both sides by 3 yields:

$a = -4$

This value of $a = -4$ is consistent with our earlier deduction that $a < 0$.

Step 3: Define the function g(x)

Now that we have found $a = -4$, we can define $g(x)$:

$g(x) = ax^2 - 6x + 15 = -4x^2 - 6x + 15$

Step 4: Analyze g(x) to determine its local extremum

We need to determine whether $g(x)$ has a local maximum or minimum and where it occurs. Since the leading coefficient of $g(x)$ is $-4$, which is negative, the parabola opens downwards, and $g(x)$ has a local maximum.

The x-coordinate of the vertex of $g(x)$ is given by $x_v = -\frac{B}{2A}$. In this case, $A = -4$ and $B = -6$. Therefore:

$x_v = -\frac{-6}{2(-4)} = -\frac{6}{8} = -\frac{3}{4}$

Thus, $g(x)$ has a local maximum at $x = -\frac{3}{4}$.

Common Mistakes & Tips

• Sign Errors: Double-check the signs when applying the vertex formula, especially when $B$ is negative.
• Confusing f(x) and g(x): Remember that the coefficient of the linear term (x term) changes from +6 in f(x) to -6 in g(x).
• Interpreting the Leading Coefficient: A negative leading coefficient implies a local maximum, while a positive leading coefficient implies a local minimum.

Summary

We first used the given information about the monotonicity of $f(x)$ to determine the value of 'a'. Knowing that the vertex of $f(x)$ occurs at $x = \frac{3}{4}$, we used the vertex formula to find $a = -4$. Then, we substituted this value into the expression for $g(x)$ and used the vertex formula again to find that $g(x)$ has a local maximum at $x = -\frac{3}{4}$.

The final answer is \boxed{local maximum at x = -3/4}, which corresponds to option (A).