Question

a parabola opening up or down has vertex (0, -4) and passes through (12, 14). 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 vertex \((0, -4)\), so \( h = 0 \) and \( k = -4 \). Substituting these values, the equation becomes \( y = a(x - 0)^2 - 4 \), which simplifies to \( y = ax^2 - 4 \).

Step2: Substitute the point \((12, 14)\)

We know the parabola passes through \((12, 14)\), so substitute \( x = 12 \) and \( y = 14 \) into \( y = ax^2 - 4 \). We get \( 14 = a(12)^2 - 4 \).

Step3: Solve for \( a \)

First, calculate \( 12^2 = 144 \), so the equation is \( 14 = 144a - 4 \). Add 4 to both sides: \( 14 + 4 = 144a \), which is \( 18 = 144a \). Then divide both sides by 144: \( a=\frac{18}{144}=\frac{1}{8} \).

Step4: Write the final equation

Substitute \( a = \frac{1}{8} \), \( h = 0 \), and \( k = -4 \) back into the vertex form \( y = a(x - h)^2 + k \). We get \( y=\frac{1}{8}(x - 0)^2 - 4 \), which simplifies to \( y=\frac{1}{8}x^2 - 4 \).

Answer:

\( y = \frac{1}{8}x^2 - 4 \)