Answer :
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
### (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
