How do you graph #y=-x^2+2x+4#?

1 Answer
Jul 14, 2015

Substitute values for #x#, solve for #y#, draw a graph through the points. Be sure to use positive and negative values for #x#.

Explanation:

#y=-x^2+2x+4#

Point A: #(-3,-11)#

#x=-3#
#y=-(3^2)+(2*-3)+4=-9-6+4=-11#

Point B: #(-1,1)#

#x=-1#
#y=-(-1^2)+(2*-1)+4=-1-2+4=1#

Point C: #(0,4)#

#x=0#
#y=-(0^2)+(2*0)+4=4#

Point D: #(1,5)#

#x=1#
#y=-(1^2)+(2*1)+4=-1+2+4=5#

Point E: #(3,1)#

#x=3#
#y=-(3^2)+(2*3)+4=-9+6+4=1#

Plot the points and draw a graph through the points. It will be a parabola.

graph{y=-x^2+2x+4 [-11.72, 8.28, -3.77, 6.23]}