Let's analyze the polynomial function [tex]\( f(x) = -(x+1)^2(x-1)^2 \)[/tex] step by step to answer the questions about its graph.
### (a) End Behavior of the Graph
To determine the end behavior of [tex]\( f(x) \)[/tex], we need to look at the degree and the leading coefficient of the polynomial. Our polynomial [tex]\( f(x) = -(x+1)^2(x-1)^2 \)[/tex] can be expanded to [tex]\( f(x) = -[(x+1)(x+1)(x-1)(x-1)] \)[/tex].
- The polynomial is of degree 4 (since the highest power of [tex]\( x \)[/tex] is [tex]\( x^4 \)[/tex]).
- The leading term after expansion would involve multiplying [tex]\( x^2 \)[/tex] from [tex]\((x+1)^2\)[/tex] and [tex]\( x^2 \)[/tex] from [tex]\((x-1)^2\)[/tex], resulting in [tex]\( x^4 \)[/tex], but when multiplied by -1, it becomes [tex]\(-x^4\)[/tex].
Since the leading coefficient is negative and the degree is even (4), the end behavior is:
The graph of [tex]\( f(x) \)[/tex] falls to negative infinity as [tex]\( x \)[/tex] approaches positive or negative infinity.
Selected End Behavior:
- The graph falls to negative infinity as [tex]\( x \)[/tex] approaches positive or negative infinity.
### (b) Real Zeros and Behavior at the x-axis
Next, we identify the real zeros of [tex]\( f(x) \)[/tex] and how the graph behaves at each zero.
To find the zeros, we set the function equal to zero:
[tex]\[ -(x+1)^2(x-1)^2 = 0 \][/tex]
This implies:
[tex]\[ (x+1)^2 = 0 \quad \text{or} \quad (x-1)^2 = 0 \][/tex]
Solving these equations gives:
[tex]\[ x = -1 \quad \text{and} \quad x = 1 \][/tex]
Both of these zeros have even multiplicity (2 in each case). When a zero has an even multiplicity, the graph touches the x-axis at these zeros but does not cross it.
Zeros where the graph crosses the x-axis:
- None (since all zeros have even multiplicity)
Zeros where the graph touches, but does not cross the x-axis:
- [tex]\(-1, 1\)[/tex]
### (c) y-intercept
To find the y-intercept of the graph, we evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -(0 + 1)^2(0 - 1)^2 = -(1^2)(-1^2) = -1 \][/tex]
y-intercept:
- [tex]\(-1\)[/tex]
### (d) Graph [tex]\( f(x) = -(x+1)^2(x-1)^2 \)[/tex]
To graph [tex]\( f(x) \)[/tex], plot the points and consider the behavior determined above:
1. Plotting Points:
- Real zeros ([tex]\( x \)[/tex]-intercepts) at [tex]\( (-1, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
- [tex]\( y \)[/tex]-intercept at [tex]\( (0, -1) \)[/tex].
2. Behavior at the Real Zeros:
- At [tex]\( x = -1 \)[/tex] and [tex]\( x = 1 \)[/tex], the graph touches the x-axis but does not cross it.
3. End Behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to -\infty \)[/tex].
By connecting these key points and following the determined behaviors, you can sketch the graph of [tex]\( f(x) \)[/tex]. The graph should touch the x-axis at [tex]\( -1 \)[/tex] and [tex]\( 1 \)[/tex], cross the y-axis at [tex]\( -1 \)[/tex], and approach [tex]\(-\infty\)[/tex] as [tex]\( x \)[/tex] approaches both positive and negative infinity.
