Question

topic 3 characteristics of polynomial functions skills practice continued
9 ( g(x) = )
graph of ( f(x) ) with points (0,0), (1,1), (2,8) and graph of ( g(x) ) with points (-2,1), (-1,0), (0,-1)
10 ( g(x) = )
graph of ( f(x) ) with points (0,0), (1,1), (2,16) and graph of ( g(x) ) with points (-2,0), (-1,-0.5), (0,-8)

Explanation:

Step1: Analyze \( f(x) \) for Problem 9

For the left graph ( \( f(x) \) ), we check points. At \( x = 1 \), \( f(1)=1 \); \( x = 2 \), \( f(2)=8 \); \( x = 0 \), \( f(0)=0 \). The shape suggests \( f(x)=x^3 \) (since \( 1^3 = 1 \), \( 2^3 = 8 \), \( 0^3 = 0 \)).

Step2: Analyze \( g(x) \) for Problem 9

For the right graph ( \( g(x) \) ), let's find the transformation. Let's assume \( g(x)=af(bx)+c \). The root at \( x=-1 \), \( x = 0 \) (but \( g(0)=-1 \)). Let's test \( x=-1 \): \( g(-1)=0 \). If \( f(x)=x^3 \), let's see the transformation. Let's check \( x=-2 \): \( g(-2)=1 \). Suppose \( g(x)= -f(-x) - 1 \)? Wait, \( f(x)=x^3 \), then \( -f(-x)= -(-x)^3 = x^3 \). But \( g(0)=-1 \), so \( g(x)= -f(-x) - 1 \)? Wait, no. Wait, \( f(x)=x^3 \), let's see the points. For \( g(x) \), when \( x=-1 \), \( g(-1)=0 \); \( x=0 \), \( g(0)=-1 \); \( x=-2 \), \( g(-2)=1 \). Let's assume \( g(x)= -f(-x) - 1 \)? Wait, \( f(-x)=(-x)^3=-x^3 \), so \( -f(-x)=x^3 \), then \( x^3 - 1 \)? At \( x=-1 \), \( (-1)^3 -1 = -2
eq 0 \). Wait, maybe \( g(x)= -f(-x) + c \)? Wait, maybe \( f(x)=x^3 \), and \( g(x) \) is a transformation. Let's check the shape: \( f(x) \) is a cubic (odd function, passes through origin, increasing). \( g(x) \) is a cubic transformed: reflected, shifted. Wait, the right graph of \( g(x) \) has a root at \( x=-1 \), so let's suppose \( g(x)= -(-x + 1)^3 - 1 \)? No, better to use points. Let's take \( f(x)=x^3 \), then \( g(x)= -f(-x) - 1 \)? Wait, \( x=-1 \): \( -f(1) -1 = -1 -1 = -2
eq 0 \). Wait, maybe \( g(x)= -f(x + 1) - 1 \)? \( f(x + 1)=(x + 1)^3 \), so \( - (x + 1)^3 - 1 \). At \( x=-1 \): \( -0 -1 = -1
eq 0 \). Wait, maybe I made a mistake. Wait, the left graph: \( f(x) \) has points (0,0), (1,1), (2,8) – so \( f(x)=x^3 \). The right graph: \( g(x) \) has points (-2,1), (-1,0), (0,-1). Let's see the pattern: when \( x=-2 \), \( y=1 \); \( x=-1 \), \( y=0 \); \( x=0 \), \( y=-1 \). So the difference in \( x \) is +1, \( y \) is -1. So it's a linear change? No, it's a cubic. Wait, \( x=-2 \): \( (-2 + 1)^3 = (-1)^3 = -1 \), but \( y=1 \), so \( -(-1) = 1 \). So \( g(x)= - (x + 1)^3 \)? At \( x=-1 \), \( -0 = 0 \) (correct). At \( x=0 \), \( - (1)^3 = -1 \) (correct). At \( x=-2 \), \( - (-1)^3 = 1 \) (correct). Yes! So \( g(x)= - (x + 1)^3 \). But \( f(x)=x^3 \), so \( g(x)= -f(x + 1) \). Wait, \( f(x + 1)=(x + 1)^3 \), so \( g(x)= -f(x + 1) \).

Step3: Analyze \( f(x) \) for Problem 10

For the left graph ( \( f(x) \) ) in problem 10, at \( x=1 \), \( f(1)=1 \); \( x=2 \), \( f(2)=16 \); \( x=0 \), \( f(0)=0 \). The shape suggests \( f(x)=x^4 \)? Wait, \( 1^4 = 1 \), \( 2^4 = 16 \), \( 0^4 = 0 \). Yes, \( f(x)=x^4 \) (since it's a even function, symmetric about y-axis, and \( x^4 \) fits \( (1)^4=1 \), \( (2)^4=16 \), \( 0^4=0 \)).

Step4: Analyze \( g(x) \) for Problem 10

For the right graph ( \( g(x) \) ) in problem 10, points: \( x=-2 \), \( g(-2)=0 \); \( x=-1 \), \( g(-1)=-0.5 \); \( x=0 \), \( g(0)=-8 \). Let's assume \( f(x)=x^4 \). Let's find the transformation. Let's check \( x=-1 \): \( g(-1)=-0.5 \). If \( f(x)=x^4 \), then maybe \( g(x)= - \frac{1}{2}f(x + 2) - 0 \)? Wait, \( f(x + 2)=(x + 2)^4 \). At \( x=-2 \), \( (0)^4 = 0 \), so \( - \frac{1}{2}(0) = 0 \) (matches \( g(-2)=0 \)). At \( x=-1 \), \( (1)^4 = 1 \), so \( - \frac{1}{2}(1) = -0.5 \) (matches \( g(-1)=-0.5 \)). At \( x=0 \), \( (2)^4 = 16 \), so \( - \frac{1}{2}(16) = -8 \) (matches \( g(0)=-8 \)). So \( g(x)= - \frac{1}{2}f(x + 2) \), and \( f(x)=x^4 \), so \( g(x)= - \frac{1}{2}(x + 2)^4 \).

Answer:

(for Problem 9 and 10):

Problem 9:

\( g(x)= -(-x + 1)^3 - 1 \)? Wait, no, from step 2, we found \( g(x)= -f(x + 1) \) where \( f(x)=x^3 \), so \( g(x)= - (x + 1)^3 \). Let's verify: \( x=-1 \): \( -0 = 0 \) (correct), \( x=0 \): \( -1 \) (correct), \( x=-2 \): \( -(-1)^3 = 1 \) (correct). So \( g(x)= - (x + 1)^3 \) or \( g(x)= -f(x + 1) \) with \( f(x)=x^3 \).

Problem 10:

\( g(x)= - \frac{1}{2}(x + 2)^4 \) (since \( f(x)=x^4 \), and the transformation is vertical compression by \( \frac{1}{2} \), reflection over x-axis, horizontal shift left by 2 units). Let's verify: \( x=-2 \): \( - \frac{1}{2}(0)^4 = 0 \) (correct), \( x=-1 \): \( - \frac{1}{2}(1)^4 = -0.5 \) (correct), \( x=0 \): \( - \frac{1}{2}(2)^4 = -8 \) (correct).

(Note: The exact form may vary based on the assumed \( f(x) \), but the key is identifying the transformation from \( f(x) \) to \( g(x) \) by analyzing key points and the shape of the polynomial graphs.)