Question

Let ​ f(x)=x2+4x+12 ​ . What is the vertex form of f(x)? What is the minimum value of f(x)?

Answer 1

`\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+4}x\stackrel{\stackrel{c}{\downarrow }}{+12} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{4}{2(1)}~~,~~12-\cfrac{4^2}{4(1)} \right)\implies (-2~~,~~12-4)\implies (-2~,~8)`

well, the quadratic has a leading term with a positive coefficient, meaning is a parabola opening upwards, like a "bowl", comes from above down down down, reaches a U-turn, namely the vertex, and goes back up up up.

so the minimum value is at the vertex of course, and the minumum is well, just the y-coordinate of the vertex, 8.