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

2 Answers
Jim G.
Aug 10, 2017

#(1/2,11/2)#

Explanation:

#"given the equation of a parabola in standard form"#

#"that is "y=ax^2+bx+c#

#"then "x_(color(red)"vertex")=-b/(2a)#

#y=-2x^2+2x+5" is in standard form"#

#"with "a=-2,b=+2,c=5#

#rArrx_(color(red)"vertex")=-2/(-4)=1/2#

#"substitute this value into the equation for the corresponding"#
#"y-coordinate"#

<#rArry_(color(red)"vertex")=-2(1/2)^2+2(1/2)+5=11/2#

#rArrcolor(magenta)"vertex "=(1/2,11/2)#

Jayden N.
Aug 10, 2017

Vertex is at #(1/2, 11/2)#.

Explanation:

The axis of symmetry is also the x value of the vertex. So we can use the formula #x=(-b)/(2a)# to find the axis of symmetry.

#x=(-(2))/(2(-2))#

#x=1/2#

Substitute #x=1/2# back into the original equation for the y value.

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

#y = 11/2#

Therefore, the vertex is at #(1/2, 11/2)#.

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