Question

identify the graph of $p(x) = x^5 - 3x^3 + 2$

Explanation:

Step1: Analyze the leading term

The function is \( p(x) = x^5 - 3x^3 + 2 \). The leading term is \( x^5 \), which has an odd degree and positive leading coefficient. So as \( x \to +\infty \), \( p(x) \to +\infty \), and as \( x \to -\infty \), \( p(x) \to -\infty \). This eliminates graphs where the end - behaviors don't match (e.g., if a graph has both ends going up or both going down).

Step2: Evaluate at \( x = 0 \)

Substitute \( x = 0 \) into the function: \( p(0)=0^5 - 3\times0^3+2 = 2 \). So the y - intercept is \( (0,2) \). We can eliminate graphs where the y - intercept is not 2.

Step3: Analyze critical points (optional but helpful)

First, find the derivative \( p^\prime(x)=5x^4 - 9x^2=x^2(5x^2 - 9)=x^2(\sqrt{5}x - 3)(\sqrt{5}x + 3) \). The critical points are at \( x = 0,x=\pm\frac{3}{\sqrt{5}}\approx\pm1.34 \). The function will have local maxima and minima around these points. But from the y - intercept and end - behavior, we can also check the value at \( x = 2 \): \( p(2)=2^5-3\times2^3 + 2=32-24 + 2 = 10 \), and at \( x=-2 \): \( p(-2)=(-2)^5-3\times(-2)^3 + 2=-32 + 24+2=-6 \).
Looking at the graphs, the graph with y - intercept 2, end - behavior (down as \( x\to-\infty \), up as \( x\to+\infty \)) and passing through appropriate points (like \( x = 2 \) giving a positive value and \( x=-2 \) giving a negative value) is the bottom - right graph (the fourth graph, the one with the blue curve that has a y - intercept of 2, goes down as \( x\to-\infty \), up as \( x\to+\infty \), and has a local minimum around \( x = 2 \) and a local maximum around \( x=-2 \) - like shape).

Answer:

The bottom - right graph (the fourth graph in the given set of graphs)