Questions: Use the given function to complete parts (a) through (e) below. f(x)=x^4-1x^2 a) Use the Leading Coefficient Test to determine the graph's end behavior. A. The graph of f(x) falls left and falls right. B. The graph of f(x) rises left and rises right. C. The graph of f(x) rises left and falls right. D. The graph of f(x) falls left and rises right. b) Find the x-intercepts. x=

Solution Steps

a) To determine the end behavior of the graph using the Leading Coefficient Test, we need to look at the leading term of the polynomial function. The leading term is the term with the highest power of \(x\). In this case, the leading term is \(x^4\). Since the coefficient of \(x^4\) is positive and the degree is even, the graph of \(f(x)\) rises to the left and rises to the right.

b) To find the \(x\)-intercepts, we need to solve the equation \(f(x) = 0\). This involves solving the polynomial equation \(x^4 - x^2 = 0\).

To analyze the end behavior of the function \( f(x) = x^4 - x^2 \), we identify the leading term, which is \( x^4 \). Since the leading coefficient is positive and the degree of the polynomial is even, we conclude that the graph of \( f(x) \) rises to the left and rises to the right.

To find the \( x \)-intercepts, we set the function equal to zero: \[ f(x) = x^4 - x^2 = 0 \] Factoring the equation, we have: \[ x^2(x^2 - 1) = 0 \] This gives us: \[ x^2 = 0 \quad \text{or} \quad x^2 - 1 = 0 \] From \( x^2 = 0 \), we find \( x = 0 \). From \( x^2 - 1 = 0 \), we find \( x = \pm 1 \). Therefore, the \( x \)-intercepts are: \[ x = -1, 0, 1 \]

Final Answer