Question

Find the vertex of the parabola \( 4y = x^2 - 6x + 17 \).

• A. \( (3, 7) \)
• B. \( (1, 4) \)
• C. \( (3, 4) \)
• D. \( (1, 7) \)

Hint:

To find the vertex of a parabola in the form \( ax^2 + bx + c \), complete the square to rewrite the equation in the form \( a(x - h)^2 + k \), where \( (h, k) \) is the vertex.

Answer 1

The Correct Option is C

The given equation of the parabola is: \[ 4y = x^2 - 6x + 17 \] We need to find the vertex of the parabola.

1. Step 1: Rewrite the equation in standard form. First, divide the entire equation by 4: \[ y = \frac{1}{4}(x^2 - 6x) + \frac{17}{4} \] Now, complete the square for the \( x \)-terms.

2. Step 2: Complete the square. To complete the square on \( x^2 - 6x \), take half of -6, which is -3, square it to get 9. Add and subtract 9 inside the bracket: \[ y = \frac{1}{4}[(x^2 - 6x + 9) - 9] + \frac{17}{4} \] Simplifying: \[ y = \frac{1}{4}[(x - 3)^2 - 9] + \frac{17}{4} \] Distribute the \( \frac{1}{4} \): \[ y = \frac{1}{4}(x - 3)^2 - \frac{9}{4} + \frac{17}{4} \] Simplify the constants: \[ y = \frac{1}{4}(x - 3)^2 + 2 \]

3. Step 3: Identify the vertex. The equation is now in the form \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. From the equation, we see that the vertex is at \( (3, 4) \). Thus, the vertex of the parabola is \( (3, 4) \).