Question

question 10 of 25
find the equation of the parabola oriented with a vertical axis of symmetry whose vertex is (3, 6) and which passes through the point (4, 4).
\\( y = -2(x - 3)^2 + 6 \\)
\\( y = 2(x - 3)^2 + 6 \\)
\\( y = -2(x + 3)^2 - 6 \\)
\\( y = (x + 3)^2 - 6 \\)

Explanation:

Step1: Recall vertex form of parabola

The vertex form of a parabola with vertical axis of symmetry is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Here, vertex is \((3, 6)\), so \( h = 3 \), \( k = 6 \). Thus, the equation becomes \( y = a(x - 3)^2 + 6 \).

Step2: Substitute the point \((4, 4)\)

Substitute \( x = 4 \), \( y = 4 \) into \( y = a(x - 3)^2 + 6 \):
\( 4 = a(4 - 3)^2 + 6 \)
Simplify: \( 4 = a(1)^2 + 6 \) → \( 4 = a + 6 \).
Solve for \( a \): \( a = 4 - 6 = -2 \).

Step3: Write the final equation

Substitute \( a = -2 \), \( h = 3 \), \( k = 6 \) into vertex form: \( y = -2(x - 3)^2 + 6 \).

Answer:

\( y = -2(x - 3)^2 + 6 \) (corresponding to the first option)