Question

The vertex of the parabola \( y^2 - 4y - x + 3 = 0 \) is:

• A. \( (-1, 3) \)
• B. \( (-1, 2) \)
• C. \( (2, -1) \)
• D. \( (3, -1) \)
• E. \( (1, 2) \)

Hint:

The \( y \)-coordinate of the vertex for \( y^2 + By + Cx + D = 0 \) is always \( -B/2 \). Here, \( B = -4 \), so the \( y \)-coordinate is \( -(-4)/2 = 2 \). Substituting \( y = 2 \) into the original equation gives \( 4 - 8 - x + 3 = 0 \implies -x - 1 = 0 \implies x = -1 \).

Answer 1

The Correct Option is B

Concept: The vertex form of a horizontal parabola is \( (y - k)^2 = 4a(x - h) \), where the vertex is located at the coordinates \( (h, k) \). To find the vertex of the given equation, we must complete the square for the \( y \)-terms.

Step 1: Group \( y \)-terms and move others to the right.
\[ y^2 - 4y = x - 3 \]

Step 2: Complete the square for \( y \).
Add \( (\frac{-4}{2})^2 = 4 \) to both sides: \[ y^2 - 4y + 4 = x - 3 + 4 \] \[ (y - 2)^2 = x + 1 \] \[ (y - 2)^2 = 1(x - (-1)) \]

Step 3: Identify the vertex.
Comparing this with \( (y - k)^2 = 4a(x - h) \), we find \( h = -1 \) and \( k = 2 \). The vertex is \( (-1, 2) \).