What is the vertex form of #y=-x^2+5x#?

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Answer 1 · 2017-12-21T00:58:50

#(x - 5/2)^2 - 25/4#

To find the vertex form, you need to complete the square:

#-x^2 + 5x#
#= x^2 - 5x#
#= x^2 - 5x + (5/2)^2 - (5/2)^2#
#= (x - 5/2)^2 - (5/2)^2#
#= (x - 5/2)^2 - 25/4#

Answer 2 · 2017-12-21T01:16:30

#y=-(x-5/2)^2+25/4#

Given -

#y=-x^2+5x#

Vertex

#x=(-b)/(2a)=(-5)/(-1xx2)=5/2#

At #x=5/2#;

#y=-(5/2)^2+5(5/2)=-25/4+25/2=(-25+50)/4=25/4#

Vertex #(5/2, 25/4)#

The vertex form of the quadratic equation is -

#y=a(x-h)^2+k#

Where -

#a=-1# - coefficient of #x^2#
#h=5/2# - x - coordinate of the vertex
#k=25/4# - y - coordinate of the vertex

Substitute these values in the formula

#y=-1(x-5/2)^2+25/4#

#y=-(x-5/2)^2+25/4#