Question 1117498
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The solution by the other tutor, using x = -b/2a for the x coordinate of the vertex, is one good way to work a problem like this.<br>
Another useful method to know is to put the equation in standard form:<br>
{{{y-k = a(x-h)^2}}}<br>
or<br>
{{{y = a(x-h)^2+k}}}<br>
In either of these forms, the vertex is at (h,k).<br>
For your example....<br>
{{{y = -x^2+ 2x +3}}}<br>
factor out the leading coefficient<br>
{{{y = -1(x^2-2x)+3}}}<br>
complete the square<br>
{{{y = -1(x^2-2x+1)+3+1}}}<br>
{{{y = -1(x-1)^2+4}}} or {{{y-4 = -1(x-1)^2}}}<br>
These forms tell you that the vertex is at (1,4); the a=-1 tells you that the parabola opens downward.  That makes the maximum value of the function is 4; so the range is y <= 4.