## Precalculus (10th Edition)

The function has a minimum value of $-21$.
RECALL: (1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens: (i) upward when $a \gt0$ and has its vertex as its minimum; or (ii) downward when $a\lt0$ and has its vertex as its maximum. (2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$ Compare $f(x)=2x^2+12x-3$ to $f(x)=ax^2+bx+c$. It can be seen that $a=2$, $b=12$, $c=-3$. $a\gt0$, hence the graph opens up, hence it's vertex is a minimum. The minimum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot 2}=-3.$ Thus, the minimum value is $f(-3)=2\cdot(-3)^2+12\cdot(-3)-3=-21.$