Question

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

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

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

Step3: Solve for \(a\)

Simplify the right - hand side: \((-4)^2=16\), so the equation is \(-4 = 16a\). To solve for \(a\), divide both sides by 16: \(a=\frac{-4}{16}\). Simplify the fraction: \(a =-\frac{1}{4}\).

Step4: Write the final equation

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

Answer:

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