## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$f(x)=-(x+3)^2+5$
The vertex form of the quadratic function $ax^2+bx+c=0$ can be expressed as $f(x)=a(x-h)^2+k$ and its vertex is at $(h, k)$. As depicted in the picture, the vertex of the graph is at $(h, k)=(-3,5)$, and thus, the quadratic function becomes $f(x)=a(x +3)^2+5$. Plug in the values $(0, -4)$ to obtain: $-4=a(0+3)^2 + 5 \\ 9a=-9 \implies a=-1$ Therefore, the equation of the function can be expressed as: $f(x)=-(x+3)^2+5$.