Question 924886
 <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi
f(x)= ax^2 + bx + c
f(x) = -x^2+2x+5    P(0,5) on this parabola
the vertex form of a Parabola opening up(a>0) or down(a<0), {{{y=a(x-h)^2 +k}}}
f(x) = -x^2+2x+5 
Completing the Square to Find Vertex Form  
f(x) = -(x - 1)^2 + + 1 + 5     Note: h = {{{-(2/2(-1)) }}} = 1 
f(x) = -(x - 1)^2 + 6
parabola opening downward a = -1 < 0
P(0,5), V(1, 6) <u>Line of symmetry</u>,   x = 1
f(x) = 0,  x = 1 ± &#8730;6
Sketch
{{{drawing(300,300,   -6, 6, -6, 10, grid(1), blue(line(1,10,1,-6)) ,
circle(0, 5,0.3),
circle(1,6,0.3),
circle(1 + sqrt(6),0,0.3),
circle(1 - sqrt(6),0,0.3),
graph( 300, 300, -6, 6, -6, 10,0,  -x^2+2x+5 ) )}}}