What is the vertex of y=-2x^2 + 2x + 5 y=−2x2+2x+5?

2 Answers
Jim G.
Aug 10, 2017

(1/2,11/2)(12,112)

Explanation:

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

"that is "y=ax^2+bx+cthat is y=ax2+bx+c

"then "x_(color(red)"vertex")=-b/(2a)then xvertex=−b2a

y=-2x^2+2x+5" is in standard form"y=−2x2+2x+5 is in standard form

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

rArrx_(color(red)"vertex")=-2/(-4)=1/2⇒xvertex=−2−4=12

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

<rArry_(color(red)"vertex")=-2(1/2)^2+2(1/2)+5=11/2⇒yvertex=−2(12)2+2(12)+5=112

rArrcolor(magenta)"vertex "=(1/2,11/2)⇒vertex =(12,112)

Jayden N.
Aug 10, 2017

Vertex is at (1/2, 11/2)(12,112).

Explanation:

The axis of symmetry is also the x value of the vertex. So we can use the formula x=(-b)/(2a)x=−b2a to find the axis of symmetry.

x=(-(2))/(2(-2))x=−(2)2(−2)

x=1/2x=12

Substitute x=1/2x=12 back into the original equation for the y value.

y = -2(1/2)^2 + 2(1/2) + 5y=−2(12)2+2(12)+5

y = 11/2y=112

Therefore, the vertex is at (1/2, 11/2)(12,112).

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