Question

a parabola opening up or down has vertex (0, 3) and passes through (6, 6). write its equation in vertex form. simplify any fractions.

Explanation:

Step1: Recall vertex form of parabola

The vertex form of a parabola opening up or down is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Given the vertex \((0, 3)\), so \( h = 0 \) and \( k = 3 \). Substituting these values, the equation becomes \( y = a(x - 0)^2 + 3 \), which simplifies to \( y = ax^2 + 3 \).

Step2: Substitute the point \((6, 6)\) into the equation

We know the parabola passes through \((6, 6)\), so substitute \( x = 6 \) and \( y = 6 \) into \( y = ax^2 + 3 \). We get \( 6 = a(6)^2 + 3 \).

Step3: Solve for \( a \)

First, simplify the equation: \( 6 = 36a + 3 \). Subtract 3 from both sides: \( 6 - 3 = 36a \), which gives \( 3 = 36a \). Then divide both sides by 36: \( a=\frac{3}{36}=\frac{1}{12} \).

Step4: Write the final equation

Substitute \( a = \frac{1}{12} \), \( h = 0 \), and \( k = 3 \) back into the vertex form \( y = a(x - h)^2 + k \). We get \( y=\frac{1}{12}(x - 0)^2 + 3 \), which simplifies to \( y=\frac{1}{12}x^2 + 3 \).

Answer:

\( y = \frac{1}{12}x^2 + 3 \)