Question

The vertex of the parabola $y=ax^2+bx+c$ is

• A. $(\frac{b}{2a}, \frac{b^2-4ac}{4a})$
• B. $(\frac{b}{2a}, \frac{4ac-b^2}{4a})$
• C. $(\frac{-b}{2a}, \frac{b^2-4ac}{4a})$
• D. $(\frac{-b}{2a}, \frac{4ac-b^2}{4a})$

Hint:

Vertex $x$-coordinate is always $-b/2a$ for any quadratic function.

Answer 1

The Correct Option is D

Step 1: Concept The vertex of a vertical parabola $y = ax^2 + bx + c$ occurs at $x = -b/2a$.

Step 2: Meaning The $y$-coordinate is found by substituting $x = -b/2a$ back into the quadratic equation.

Step 3: Analysis $y = a(-b/2a)^2 + b(-b/2a) + c = ab^2/4a^2 - b^2/2a + c = b^2/4a - 2b^2/4a + 4ac/4a$.

Step 4: Conclusion Simplifying gives $y = (4ac - b^2)/4a$. Thus vertex is $(-b/2a, (4ac-b^2)/4a)$. Final Answer: (D)