SOLUTION: Need help graphing f(x)=-x^2+2x+5

 

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), 

f(x) = -x^2+2x+5 

Completing the Square to Find Vertex Form  

f(x) = -(x - 1)^2 + + 1 + 5     Note: h =  = 1 

f(x) = -(x - 1)^2 + 6

parabola opening downward a = -1 < 0

P(0,5), V(1, 6) Line of symmetry,   x = 1

f(x) = 0,  x = 1 ± √6

Sketch