Question

URGENT!!!

Consider the graph of the polynomial function y=f(x) given below. The graph continues forever at both ends. Determine the following features (make sure you write all intervals in interval notation and all intercepts as points):

(A) State the domain and range.

(B) State the x-intercepts.

(C) State the y-intercept.

(D) State the interval(s) for which the function is increasing.

(E) State the interval(s) for which the function is decreasing.

(F) State the interval(s) for which the function is constant.

(G) State any symmetry about any axis and/or the origin.

(H) Determine whether the function is even, odd, or neither.

Answer 1

(A)
Domain:
`\mathbf {(-\infty, \infty)}`
Range :
`\mathbf {(-\infty, \infty)}`

(B)

`(-2√(2) , 0) \;and\; (2√(2) , 0)`

(C)
(- ∞, -2] and [ 0, 2 ]

(D)

(- ∞, -2] and [ 0, 2 ]

(E)

[-2, 0 ] and [2, ∞ )

(F)

There is no interval where the function is constant

(G)

The graph is symmetric about the y-axis

(H)
This is an even function

Explanation:

Warning!!
Pay careful attention to the use of two different brackets when using interval notation. A square bracket means the value next to it is included in that interval, a regular bracket indicates it is excluded

For example: [2, 4) means 2 is included but not 4

Pay careful attention to which bracket is being used where in my explanation and answers. You could lose points if you use the wrong brackets

(A)

Domain is the set of all x values that yield a real and defined value for y

Since the graph continues forever, the range of x values is -∞ to +∞

This is written as - ∞ < x < ∞ or, in interval notation as
`\mathbf {(-\infty, \infty)}`

Range is the set of y values for the given domain. Since the graph goes on forever, the range of y is - ∞ < y < ∞. Interval notation:
`\mathbf {(-\infty, \infty)}`

(B)

The x-intercepts are where the function graph crosses the x-axis. Mathematically, it is the value of x at y =

There are two x-intercepts:


`(-2√(2) , 0) \;and\; (2√(2) , 0)`

(C)
y-intercept is where the graph intersects the y-axis. Here the graph touches the y-axis at (0, 0).So y-intercept = (0,0)

(D)
The function is increasing in the intervals (- ∞, -2] and [ 0, 2 ]

(E)
The function decreases in the following intervals for values of x
[-2, 0 ] and [2, ∞ )

(F)

There is no interval where the function is constant

(G)
The graph is symmetric about the y-axis as you see visually. Mathematically, a graph is symmetric about the y axis if f(-x) = f(x)
Here we see that f(2) = 16 and f(-2) = 16 so there is y-axis symmetry

There is no symmetry about the x axis

To see if there is symmetry about the origin, replace any(x, y) with (-x, -y) and see if it is valid. (2, 16) replaced with (-2, -16) is not a point on the graph hence no symmetry about origin

(H)
This is an even function since it is symmetric about the y-axis