What are the #y# and #x# intercept(s) of #y=2x^2-4#?

3 Answers
Gió
Sep 29, 2016

We can set alternately #x=0# and #y=0# to find the intercepts:

Explanation:

To find the y-intercept set #x=0# into your expression and get:
#y=2*0-4=-4#
Sothe coordinates of the y-intercept will be:
#x=0 and y=-4#

To find the x-intercept(s) set #y=0# to get:
#2x^2-4=0#
Rearranging:
#x^2=4/2#
#x^2=2#
#x=+-sqrt(2)#
We have two intercepts of coordinates:
#x=sqrt(2) and y=0#
#x=-sqrt(2) and y=0#

Graphically we can "see" them:
graph{2x^2-4 [-8.625, 11.375, -6.64, 3.36]}

Alan P.
Sep 29, 2016

y-intercept: #y=-4#
x-intercepts: #x=-sqrt(2) and x=sqrt(2)#

Explanation:

The y-intercept is the value of #y# when #x=0#
#color(white)("XXX")y=2x^2-4# with #x=0# becomes
#color(white)("XXX")y=2 * 0^2-4 = -4#

The x-intercepts are the values of #x# when #y=0#
#color(white)("XXX")y=2x^2-4# when #y=0# becomes
#color(white)("XXX")0=2x^2-4#
#color(white)("XXX")2x^2=4#
#color(white)("XXX")x^2=2#
#color(white)("XXX")x=+_sqrt(2)#

Alan N.
Sep 29, 2016

#y# intercept #-4#, #x# intercepts #+-sqrt2#

Explanation:

#y=2x^2-4#

The #y# intercept is at #x=0#
Where: #y=-4#

The #x# intercept ia at #y=0#
Where: #2x^2-4=0#

#x^2=4/2#

#x=+-sqrt2#

These can be seen on the graph of #2x^2-4# below
graph{2x^2-4 [-6.1, 6.384, -5.12, 1.126]}

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