How do you find the vertex and the intercepts for #y= -2x^2 + 2x +2#?

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Answer 1 · 2018-04-20T12:53:18

Vertex is at #(0.5,2.5)#, y intercept is #y=2 or (0,2) # and
x intercepts are at #(-0.618,0) and (1.618,0)#

#y =-2 x^2+2 x+2 ; a= -2 , b= 2 ,c =2#

Vertex ( x coordinate) is # v_x= -b/(2 a)=(-2)/-4=1/2=0.5 #

Putting #x=2# in the equation we get #v_y#

Vertex ( y coordinate) is # v_y= -2 *1/4+2*1/2+2#

#= -1/2 + 1+2= 2.5 :. # Vertex is at #(0.5,2.5)#

y intercept is found by putting #x=0# in the equation

#y = -2 x^2+2 x+2:. y=2 ;# y intercept is #y=2 or (0,2)#

x intercept is found by putting #y=0# in the equation

#y = -2 x^2+2 x+2 or -2 x^2+2 x+ 2=0 # or

# -2( x^2+ x) = -2 or -2 ( x+0.5)^2 = -2- 0.5#

#-2(x-0.5)^2=-2.5 or (x-0.5)^2 = 1.25 # or

# (x-0.5) = +- sqrt 1.25 :. x = 0.5 +- 1.118#

#:. x ~~ 1.618 and x ~~ -0.618#

x intercepts are at #(-0.618,0) and (1.618,0)#

graph{-2x^2+2x+2 [-10, 10, -5, 5]} [Ans]