Question

a parabola opening up or down has vertex (0, 6) and passes through (-6, -3). 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 \((0,6)\), so \( h = 0 \), \( k = 6 \). Substituting these into the formula, we get \( y = a(x - 0)^2 + 6 \), which simplifies to \( y = ax^2 + 6 \).

Step2: Find the value of \( a \)

The parabola passes through the point \((-6, -3)\). This means when \( x = -6 \), \( y = -3 \). Substitute \( x = -6 \) and \( y = -3 \) into the equation \( y = ax^2 + 6 \):

$$ -3 = a(-6)^2 + 6 $$

First, calculate \((-6)^2 = 36\), so the equation becomes:

$$ -3 = 36a + 6 $$

Subtract 6 from both sides:

$$ -3 - 6 = 36a $$

$$ -9 = 36a $$

Divide both sides by 36:

$$ a = \frac{-9}{36} = -\frac{1}{4} $$

Step3: Write the final equation

Substitute \( a = -\frac{1}{4} \), \( h = 0 \), and \( k = 6 \) back into the vertex form \( y = a(x - h)^2 + k \):

$$ y = -\frac{1}{4}(x - 0)^2 + 6 $$

Simplifying, we get:

$$ y = -\frac{1}{4}x^2 + 6 $$

Answer:

\( y = -\frac{1}{4}x^2 + 6 \) (or in vertex form as \( y = -\frac{1}{4}(x - 0)^2 + 6 \))