Question

1. give the equation of a quadratic function, in vertex form, for a parabola that has a vertex at (3,16) and has a y - intercept at - 2. the general form of a quadratic in vertex form is: $y = a(x - h)^2 + k$

Explanation:

Step1: Plug vertex into vertex form

The vertex \((h,k)\) is \((3,16)\), so substitute \(h=3\), \(k=16\) into \(y=a(x-h)^2+k\):
\(y=a(x-3)^2+16\)

Step2: Use y-intercept to find \(a\)

The y-intercept is \((0,-2)\), substitute \(x=0\), \(y=-2\):
\(-2=a(0-3)^2+16\)
Simplify: \(-2=9a+16\)
Solve for \(a\): \(9a=-2-16=-18\), so \(a=\frac{-18}{9}=-2\)

Step3: Substitute \(a\) back into the equation

\(y=-2(x-3)^2+16\)

Answer:

\(y=-2(x-3)^2+16\)