How do you find the vertex and the intercepts for #f(x)=-2x^2+2x+4#?

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Answer 1 · 2017-11-25T06:50:17

Vertex #(1/2, 9/2)#
Y- intercept #(0, 4)#
X - intercept #(-1,0)#
X - intercept #(2,0)#

Given -

#f(x)=-2x^2+2x+4#

#y=-2x^2+2x+4#

Vertex

#x=(-b)/(2a)=(-2)/(2xx(-2))=(-2)/(-4)=1/2#

At #x=1/2#

#y=-2xx(1/2)^2+(2xx1/2)+4#

#y=-2xx(1/4)+1+4#

#y=--1/2+1+4=(-1+2+8)/2=9/2#

Vertex #(1/2, 9/2)#

Y- intercept
At #x=0#

#y=-2(0)^2+2(0)+4#

#y=4#

Y- intercept #(0, 4)#

X- intercept

At #y=0#

#-2x^2+2x+4=0#

#-2x^2+4x-2x+4=0#

#-2x(x-2)-2(x-2)=0#

#(-2x-2)(x-2)=0#

#-2x-2=0#
#x=(2)/(-2)=-1#

X - intercept #(-1,0)#

#x-2=0#

#x=2#

X - intercept #(2,0)#