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Use the given function to complete parts (a) through (e) below.
f(x) = x²-1x²
a) Use the Leading Coefficient Test to determine the graph's end behavior.
OA. The graph of f(x) rises left and falls right.
B. The graph of f(x) falls left and rises right.
c. The graph of f(x) rises left and rises right.
OD. The graph of f(x) falls left and falls right.
b) Find the x-intercepts.
x= -1,0,1
(Type an integer or a decimal. Use a comma to separate answers as needed.)
At which zeros does the graph of the function cross the x-axis? Select the correct
choice below and, if necessary, fill in the answer box to complete your choice.
A.
x= 1,1 (Type an integer or a decimal. Use a comma to separate answers
as needed.)
OB. There are no x-intercepts at which the graph crosses the x-axis.
At which zeros does the graph of the function touch the x-axis and turn around? Select
the correct choice below and, if necessary, fill in the answer box to complete your
choice.
(Type an integer or a decimal. Use a comma to separate answers
as needed.)
X =
○ A.
○ B.
There are no x-intercepts at which the graph touches the x-axis and turns
around.

Sagot :

GinnyAnswer
To solve the problem systematically, let's analyze and answer each part step by step:

### (a) Use the Leading Coefficient Test to determine the graph's end behavior.

Given the function [tex]\( f(x) = x^2 - 1x^2 \)[/tex], we can simplify it as follows:
[tex]\[ f(x) = x^2 - x^2 \][/tex]
[tex]\[ f(x) = 0 \][/tex]

However, assuming a typographical error in the function, leading to [tex]\( f(x) = -x^2 \)[/tex], the leading coefficient (the coefficient of [tex]\( x^2 \)[/tex]) is [tex]\(-1\)[/tex], which is negative. Additionally, since the degree (highest exponent of [tex]\( x \)[/tex]) is 2, which is even, we can use the Leading Coefficient Test:

- If the leading coefficient is negative (and the degree is even), the graph falls left and falls right.

Thus, the correct answer is:
D. The graph of f(x) falls left and falls right.

### (b) Find the x-intercepts.

To find the x-intercepts, we need to set the function equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ \text{Given function:} \quad f(x) = -x^2 \][/tex]
[tex]\[ -x^2 = 0 \][/tex]
[tex]\[ x^2 = 0 \][/tex]
[tex]\[ x = 0 \][/tex]

So, the x-intercepts are:
[tex]\[ x = [0] \][/tex]

### At which zeros does the graph of the function cross the x-axis?

The graph will cross the x-axis at those x-intercepts where the graph moves from one side of the x-axis to the other. Given the nature of quadratic functions and specifically the behavior of [tex]\( -x^2 \)[/tex], the graph touches the x-axis and turns around at [tex]\( x = 0 \)[/tex] but does not actually cross it.

So, the correct option is:
B. There are no x-intercepts at which the graph crosses the x-axis.

### At which zeros does the graph of the function touch the x-axis and turn around?

From the calculation above, at [tex]\( x = 0 \)[/tex], the graph touches the x-axis and turns around since it is a parabola opening downwards.

Thus, the correct option is:
A. x = 0
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