Question

a parabola opening up or down has vertex (0, 4) and passes through (-8, -4). 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 \( h = 0 \), \( k = 4 \). Substituting these values, the equation becomes \( y = a(x - 0)^2 + 4 \), which simplifies to \( y = ax^2 + 4 \).

Step2: Substitute the point \((-8, -4)\)

We know the parabola passes through \((-8, -4)\), so substitute \( x = -8 \) and \( y = -4 \) into \( y = ax^2 + 4 \). We get \(-4 = a(-8)^2 + 4\).

Step3: Solve for \( a \)

First, calculate \((-8)^2 = 64\). So the equation is \(-4 = 64a + 4\). Subtract 4 from both sides: \(-4 - 4 = 64a\), which gives \(-8 = 64a\). Then divide both sides by 64: \( a = \frac{-8}{64} = -\frac{1}{8} \).

Step4: Write the final equation

Substitute \( a = -\frac{1}{8} \) back into \( y = ax^2 + 4 \). So the equation in vertex form is \( y = -\frac{1}{8}x^2 + 4 \) (or \( y = -\frac{1}{8}(x - 0)^2 + 4 \)).

Answer:

\( y = -\frac{1}{8}x^2 + 4 \) (or \( y = -\frac{1}{8}(x - 0)^2 + 4 \))