How to find the x and y-intercept given #y = 1/2x^2 - 2#?

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Answer 1 · 2015-05-10T13:38:45

This parabola intersects the #y# axis at #(0, -2)# and the #x# axis at #(-2, 0)# and #(2, 0)#.

To find these intersections, start by substituting #x = 0# and #y = 0# into the equation of the curve.

First the #y# axis: Substitute #x = 0# ...

#y = (1/2)x^2 - 2 = (1/2)0^2 - 2 = 0 - 2 = -2#.

Now the #x# axis: Substitute #y = 0# to get:

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

Add 2 to both sides:

#(1/2)x^2 = 2#

Multiply both sides by 2 to get:

#x^2 = 4#

Hence #x = +-2#