Question

identify the graph of the following function, given that a > 0. f(x) = x²(x + a)

Explanation:

Step1: Analyze the function's roots

The function is \( f(x) = x^2(x + a) \). To find the roots, set \( f(x) = 0 \). So \( x^2 = 0 \) gives a root at \( x = 0 \) (with multiplicity 2), and \( x + a = 0 \) gives a root at \( x = -a \) (since \( a>0 \), \( -a < 0 \)).

Step2: Analyze the behavior at roots

• At \( x = 0 \) (multiplicity 2), the graph touches the x - axis and turns around (because even multiplicity).
• At \( x=-a \) (multiplicity 1), the graph crosses the x - axis.

Step3: Analyze the end - behavior

The leading term of \( f(x)=x^2(x + a)=x^3+ax^2 \) is \( x^3 \). For the leading term \( y = x^3 \), as \( x
ightarrow+\infty \), \( y
ightarrow+\infty \) and as \( x
ightarrow-\infty \), \( y
ightarrow-\infty \).

Step4: Analyze the number of turning points

The degree of the polynomial \( f(x) \) is \( 3 \) (since \( x^2\times x=x^3 \)). The maximum number of turning points of a polynomial of degree \( n \) is \( n - 1 \). So for a degree 3 polynomial, the maximum number of turning points is \( 3-1 = 2 \).

Now let's analyze the graphs:

• The first graph: Has a root at \( x = 0 \) (touches the axis) and a root at \( x=-2 \) (since \( a = 2>0 \), \( -a=-2 \)). The end - behavior and the number of turning points match. The graph touches the x - axis at \( x = 0 \) (multiplicity 2) and crosses at \( x=-2 \) (multiplicity 1). The number of turning points is 2, and the end - behavior (as \( x

ightarrow+\infty \), \( y
ightarrow+\infty \); as \( x
ightarrow-\infty \), \( y
ightarrow-\infty \)) is correct.

• The second graph: Has a root at \( x = 0 \) (but the behavior at \( x = 0 \) seems to cross? No, wait, the root at \( x = 0 \) should be a touch - point. Also, the root at \( x = 2 \), but our root should be at \( x=-a<0 \), so this is incorrect.
• The third graph: Has a root at \( x = 2 \), which is positive, but our root \( x=-a \) is negative (since \( a>0 \)), so incorrect.
• The fourth graph: The root at \( x = 0 \) is a touch - point, but the other root seems to be at \( x=-2 \)? Wait, no, the shape and the number of turning points and the root positions: Wait, the first graph has a root at \( x=-2 \) (left of origin) and \( x = 0 \) (touches). The end - behavior and turning points: Let's re - check. The function \( f(x)=x^2(x + a) \), when \( x=-a \) (negative), \( x = 0 \). The derivative \( f^\prime(x)=2x(x + a)+x^2=x(2(x + a)+x)=x(3x + 2a) \). Setting \( f^\prime(x)=0 \), we get \( x = 0 \) or \( x=-\frac{2a}{3} \). So the critical points are at \( x = 0 \) and \( x=-\frac{2a}{3} \) (negative, since \( a>0 \)). So there are two critical points, meaning two turning points. The first graph has two turning points, a root at \( x=-2 \) (so \( a = 2 \), \( -a=-2 \)) and a root at \( x = 0 \) (touches the axis). The end - behavior is correct (as \( x

ightarrow+\infty \), \( y
ightarrow+\infty \); as \( x
ightarrow-\infty \), \( y
ightarrow-\infty \)).

Answer:

The first graph (the top - left graph)