Question 1182226
<pre>
For the graph of
{{{y=ax^2+2x+c}}}
to have a maximum value, it must open downward which means that its leading
coefficient, a, must be negative.

y is its maximum value at the vertex when x=1, so we substitute 1 for x and 2
for y

{{{y=ax^2+2x+c}}}
{{{2=a(1)^2+2(1)+c}}}
{{{2=a+2+c}}}
{{{0=a+c}}}

The formula for the x coordinate of the vertex is:

{{{-b/(2a)}}}
So
{{{-2/(2a)=1}}}
{{{-2=2a}}}
{{{-1=a}}}

Substitute -1 for a in

{{{0=a+c}}}
{{{0=-1+c}}}
{{{1=c}}}

To check, substitute -1 for a and 1 for c in

{{{y=ax^2+2x+c}}}
{{{y=-1x^2+2x+1}}}
{{{y=-x^2+2x+1}}}

Then graph:

{{{drawing(400,800/3,-2,4,-1,3,graph(400,800/3,-2,4,-1,3, -x^2+2x+1),
circle(1,2,.05),locate(.8,2.3,"(1,2)") )}}} 

Edwin</pre>