The graph of this function has xintercepts at The graph of t

The graph of this function has x-intercept(s) at;

The graph of this function has y-intercept at;

Solution

f(x) = IxI - 1 if x>= -2

inorder to find x intercept put y (or f(x)) = 0

0 =IxI -1

IxI = 1

x can be either \"1\" or \"-1\"               (as I-1I = 1 and I1I =1 it always gives postive value)

x= 1 and x= -1 these \"x\" are greater then -2

so the x-intercept for x >= -2 are 1 and -1.

now we have to find y- intercept

put x = 0

y = 0 -1

y = -1

so the y -intercept is \"-1\".

now f(x) =x^2 +1 if x < -2

x-intercept put y = 0

0 = x^2 +1

x^2 = -1

since square of a real number can\'t be negative therefore no x-intercept

now we have to find y-intercept

f(x) = x^2 +1

we have to put x = 0 to find y-intercept, but we cannot put x=0 in x^2 +1 as this graph is defined for x < -2

therefore no y-intercept