What are the x and y intercepts of #y = 2x -sqrt(x^2 + 1)#?

1 Answer
Sep 27, 2016

x intercept #->x = sqrt(3)/3#
y intercept#->y = -1#

Explanation:

x intercepts:

These occur where #y = 0#:

#0 = 2x - sqrt(x^2 + 1)#

#sqrt(x^2 + 1) = 2x#

#(sqrt(x^2 + 1))^2 = (2x)^2#

#x^2 + 1 = 4x^2#

#1 = 3x^2#

#1/3 = x^2#

#x = +-1/sqrt(3) = +-sqrt(3)/3#

However, since we're dealing with a function that is in part radical, there can be no negative x-intercept, hence the only x intercept will be #x = sqrt(3)/3#.

y intercepts:

These occur where #x = 0#:

#y = 2(0) - (0^2 + 1)^(1/2)#

#y = 0 - sqrt(1)#

#y = -1#

Hopefully this helps!