Question

The vertex of the parabola $2y = -3x^2 + 48x - 200$, is

• A. $(8, -8)$
• B. $(8, -6)$
• C. $(-4, 8)$
• D. $(8, -4)$
• E. $(-4, -8)$

Hint:

Differentiating the expression \( f(x) = -3x^2 + 48x - 200 \) and setting it to zero is often the fastest way to find the $x$-coordinate of the vertex: \( -6x + 48 = 0 \implies x = 8 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The vertex of a downward/upward opening parabola given in the form \(y = ax^2 + bx + c\) is located at \(x = \frac{-b}{2a}\).
Step 2: Key Formula or Approach:
Rewrite the equation to isolate \(y\).
Find the x-coordinate of the vertex using \(x = \frac{-b}{2a}\).
Substitute \(x\) back into the equation to find the y-coordinate.
Alternatively, complete the square.
Step 3: Detailed Explanation:
Let's use the completion of squares method to be thorough.
\[ 2y = -3(x^2 - 16x) - 200 \] Take half of the x-coefficient (\(-16/2 = -8\)), square it (\(64\)), and add/subtract it inside:
\[ 2y = -3(x^2 - 16x + 64 - 64) - 200 \] \[ 2y = -3(x - 8)^2 - 3(-64) - 200 \] \[ 2y = -3(x - 8)^2 + 192 - 200 \] \[ 2y = -3(x - 8)^2 - 8 \] Divide by 2:
\[ y = -\frac{3}{2}(x - 8)^2 - 4 \] This is the standard vertex form \(y = a(x - h)^2 + k\).
The vertex \((h, k)\) is \((8, -4)\).
Step 4: Final Answer:
The vertex is (8, -4).