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

2 Answers
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)

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).