Question 182550
<font face="Times New Roman" size="+2">

The vertex of the parabola described by 


*[tex \LARGE \text{          }\math f(x) = ax^2 + bx + c]


is


*[tex \LARGE \text{          }\math \left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)]


So, calculate *[tex \Large \frac {-b}{2a}] to obtain the <i>x</i>-coordinate of the vertex, then calculate the value of your function at that <i>x</i>-coordinate value to obtain the <i>y</i>-coordinate of the vertex.


If *[tex \Large a > 0] then the parabola opens upward and the vertex is a minimum.


If *[tex \Large a < 0] then the parabola opens downward and the vertex is a maximum.


Since


*[tex \LARGE \text{          }\math f(x) = ax^2 + bx + c] where *[tex \Large \ a, b, c\ \in\ \R] is defined for all real <i>x</i>, the domain is:


*[tex \LARGE \text{          }\math \{x\ |\ x\ \in\ \R\}]


For a parabola that opens upward, the range is:


*[tex \LARGE \text{          }\math \{y\ |\ y\ \in\ \R,\ y \geq f\left(\frac{-b}{2a}\right)\}]


For a parabola that opens downward, the range is:


*[tex \LARGE \text{          }\math \{y\ |\ y\ \in\ \R,\ y \leq f\left(\frac{-b}{2a}\right)\}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>