Question

a parabola opening up or down has vertex (0, 3) and passes through (-8, -13). write its equation in vertex form. simplify any fractions.

Explanation:

Step1: Recall vertex form of parabola

The vertex form of a parabola that opens 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 into the formula, we get \( y = a(x - 0)^2 + 3 \), which simplifies to \( y = ax^2 + 3 \).

Step2: Substitute the point into the equation

The parabola passes through \((-8, -13)\). So we substitute \( x = -8 \) and \( y = -13 \) into \( y = ax^2 + 3 \). This gives us the equation: \(-13 = a(-8)^2 + 3\).

Step3: Solve for \( a \)

First, calculate \((-8)^2 = 64\). So the equation becomes \(-13 = 64a + 3\). Subtract 3 from both sides: \(-13 - 3 = 64a\), which simplifies to \(-16 = 64a\). Then divide both sides by 64: \( a=\frac{-16}{64}=\frac{-1}{4}\).

Step4: Write the final equation

Now that we have \( a = -\frac{1}{4} \), \( h = 0 \), and \( k = 3 \), substitute these back into the vertex form \( y = a(x - h)^2 + k \). We get \( y = -\frac{1}{4}(x - 0)^2 + 3 \), which simplifies to \( y = -\frac{1}{4}x^2 + 3 \).

Answer:

\( y = -\frac{1}{4}x^2 + 3 \) (or in vertex form as \( y = -\frac{1}{4}(x - 0)^2 + 3 \))