How do you find the vertex and intercepts for #y=2x^2-7#?

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Answer 1 · 2016-02-29T06:04:41

Its vertex is #(0,7)#
Its Y- intercept is also #(0,7)#
Its X-intercepts are #(1.87, 0); (-1.87,0)#

Given -

#y=y=2x^2-7#

We shall have it as

#y=2x^2+0x-7#

Vertex

#x=(-b)/(2a)=(-0)/2 xx 2=0/4=0#

At #x=0 #

#y=2(0)^2-7=7#

Its vertex is #(0,7)#

Its Y- intercept is also #(0,7)#

X intercept

At #y=0#

#2x^2-7=0#
#x^2=7/2=3.5#
#x=+-sqrt(3.5#
#x=1.87#
#x=-1.87#

Its X-intercepts are #(1.87, 0); (-1.87,0)#