How do you find the end behavior and state the possible number of x intercepts and the value of the y intercept given #y=x^2-4#?

1 Answer
Aug 1, 2018

"x intercepts" are #(-2,0) and (2,0)# , "y intercept" is #y=-4# . End behavior : Up ( As #x -> -oo , y-> oo#),
Up ( As #x -> oo , y-> oo#),

Explanation:

#y=x^2-4#. This is equation of parabola opening up since

leading coefficient is #(+)#.

x intercepts : Putting #y=0# in the equation we get,

#x^2-4=0 or (x+2)(x-2)=0 :. x= -2 and x=2# are

two x intercepts at #(-2,0) and (2,0)#.

y intercept: Putting #x=0# in the equation we get,

#y= 0-4 or y= -4 or (0,-4)# is y intercept.

The end behavior of a graph describes far left

and far right portions. Using degree of polynomial and leading

coefficient we can determine the end behaviors. Here degree of

polynomial is #2# (even) and leading coefficient is #+#.

For even degree and positive leading coefficient the graph goes

up as we go left in #2# nd quadrant and goes up as we go

right in #1# st quadrant.

End behavior : Up ( As #x -> -oo , y-> oo#),

Up ( As #x -> oo , y-> oo#),

graph{x^2-4 [-10, 10, -5, 5]} [Ans]