Algebra 2 (1st Edition)

Let's compare $f(x)=4x^2+16x-3$ to $f(x)=ax^2+bx+c$. We can see that a=4, b=16, c=-3. a>0, hence the graph opens up, and its vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{16}{2\cdot 4}=-2.$ Hence the minimum value is $f(-2)=4(-2)^2+16(-2)-3=-19.$