Question 984492
<pre>
The smallest value of a quadratic function with a positive leading coefficient
occurs at the vertex, of which the x-coordinate is {{{-b/(2a)}}}.  The minimum
value is the y-coordinate of the function when that x-coordinate is substituted
for x in the function.  

The largest value of a quadratic function with a negative leading coefficient
occurs at the vertex, of which the x-coordinate is {{{-b/(2a)}}}.  The maximum
value is the y-coordinate of the function when that x-coordinate is substituted
for x in the function.  

Instead of doing your problem for you, I'll do one exactly in every step and
detail like yours.  You can use it as a model to do yours.


f(x) = ax^2 + bx + c 
when a = 3, b = 12 and c = 127

f(x) = 3x^2 + 12x + 127

The x-coordinate of the vertex is

{{{-b/(2a)}}} = {{{-(12)/(2(3))}}} = {{{-12/6}}} = {{{-2}}}

We find the y-ccordinate by substituting -2 for x in

f(x) = 3x^2 + 12x + 127
f(2) = 3(-2)^2 + 12(-2) + 127
f(2) = 3(4) - 24 + 127
f(2) = 12 - 24 + 127
f(2) = 115

So the minimum value is 115.

Now do yours exactly the same way.

Edwin</pre>