Question

a parabola opening up or down has vertex (0, -6) and passes through (4, -2). 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 \( h = 0 \) and \( k=-6 \). Substituting these values, the equation becomes \( y = a(x - 0)^2 - 6 \), which simplifies to \( y = ax^2 - 6 \).

Step2: Find the value of \( a \)

The parabola passes through the point \((4,-2)\). Substitute \( x = 4 \) and \( y=-2 \) into the equation \( y = ax^2 - 6 \):

$$ -2=a(4)^2 - 6 $$

$$ -2 = 16a-6 $$

Add 6 to both sides:

$$ -2 + 6=16a $$

$$ 4 = 16a $$

Divide both sides by 16:

$$ a=\frac{4}{16}=\frac{1}{4} $$

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:

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