Question

unit 4
polynomials
unit 4 test: b

1. what is the end behavior of the function $f(x) = -3x^2 + 2x + 15$? circle the correct answer

a.) as $x \to +\infty, f(x) \to +\infty$ and as $x \to -\infty, f(x) \to +\infty$
b.) as $x \to +\infty, f(x) \to -\infty$ and as $x \to -\infty, f(x) \to -\infty$
c.) as $x \to +\infty, f(x) \to +\infty$ and as $x \to -\infty, f(x) \to -\infty$
d.) as $x \to +\infty, f(x) \to -\infty$ and as $x \to -\infty, f(x) \to +\infty$

1. sketch the graph of the function $f(x) = -x^2(x - 2)$.

graph with x-axis from -4 to 4 and y-axis from -4 to 4

1. describe the transformations of how the graph of $g(x)$ is related to the graph of $f(x) = x^3$.

$g(x) = -3(x - 5)^3$

1. given $f(x) = x^3$, find $g(x)$ from the description provided and graph of the transformed function. $f(x) = x^3$ is translated 3 units right and 4 units up. be sure to include 3 points on the graph.

graph with x-axis from -4 to 4 and y-axis from -4 to 4
$g(x) = $

Explanation:

Question 1

Step1: Identify the leading term

The function is \( f(x) = -3x^2 + 2x + 15 \). The leading term is \( -3x^2 \), with degree 2 (even) and leading coefficient -3 (negative).

Step2: Determine end behavior

For even - degree polynomials:

• If the leading coefficient is positive, as \( x \to +\infty \) and \( x \to -\infty \), \( f(x) \to +\infty \).
• If the leading coefficient is negative, as \( x \to +\infty \) and \( x \to -\infty \), \( f(x) \to -\infty \)? Wait, no, wait. Wait, for \( y = ax^n \), when \( n \) is even:
• If \( a>0 \), both ends go up ( \( x\to\pm\infty, y\to+\infty \) )
• If \( a < 0 \), both ends go down ( \( x\to+\infty, y\to-\infty \); \( x\to-\infty, y\to-\infty \) )? Wait, no, wait the options:

Wait the options are:
A: \( x\to+\infty,f(x)\to+\infty \); \( x\to-\infty,f(x)\to+\infty \)
B: \( x\to+\infty,f(x)\to-\infty \); \( x\to-\infty,f(x)\to-\infty \)
C: \( x\to+\infty,f(x)\to+\infty \); \( x\to-\infty,f(x)\to-\infty \)
D: \( x\to+\infty,f(x)\to-\infty \); \( x\to-\infty,f(x)\to+\infty \)

Wait, the leading term is \( -3x^2 \). The degree is 2 (even), leading coefficient - 3 (negative). So as \( x\to+\infty \), \( x^2\to+\infty \), times - 3, so \( f(x)\to-\infty \). As \( x\to-\infty \), \( x^2 = (-x)^2\to+\infty \), times - 3, so \( f(x)\to-\infty \). So the end behavior is as \( x\to+\infty,f(x)\to-\infty \) and as \( x\to-\infty,f(x)\to-\infty \), which is option B.

Step1: Analyze the function \( f(x)=-x^{2}(x - 2)=-x^{3}+2x^{2} \)

• Degree and leading term: The degree is 3 (odd), leading term \( -x^{3} \), leading coefficient - 1 (negative). So as \( x\to+\infty \), \( f(x)\to-\infty \); as \( x\to-\infty \), \( f(x)\to+\infty \) (since for \( y = ax^n \), \( n \) odd: \( a>0 \) means up on right, down on left; \( a < 0 \) means down on right, up on left).
• Zeros: Set \( f(x) = 0 \). \( -x^{2}(x - 2)=0 \) gives \( x = 0 \) (multiplicity 2) and \( x = 2 \) (multiplicity 1).
• At \( x = 0 \) (multiplicity 2, even), the graph touches the x - axis and turns around.
• At \( x = 2 \) (multiplicity 1, odd), the graph crosses the x - axis.

Step2: Find some points

• When \( x = 0 \), \( f(0)=0 \)
• When \( x = 2 \), \( f(2)=0 \)
• When \( x = 1 \), \( f(1)=-1^{2}(1 - 2)=-1\times(-1) = 1 \)
• When \( x=-1 \), \( f(-1)=-(-1)^{2}(-1 - 2)=-1\times(-3)=3 \)
• Now, plot the points: (0,0) (touches and turns), (2,0) (crosses), (1,1), (-1,3). Also, use the end - behavior: as \( x\to+\infty \), \( f(x)\to-\infty \); as \( x\to-\infty \), \( f(x)\to+\infty \). So the graph will come from the top left (since \( x\to-\infty,f(x)\to+\infty \)), pass through (-1,3), touch at (0,0), go up a bit? Wait no, wait \( f(x)=-x^{2}(x - 2)=-x^{3}+2x^{2} \). The derivative \( f^\prime(x)=-3x^{2}+4x=x(-3x + 4) \). Critical points at \( x = 0 \) and \( x=\frac{4}{3}\approx1.33 \). At \( x = 0 \), it's a local maximum (since the derivative changes from positive to negative? Wait, for \( x<0 \), say \( x=-1 \), \( f^\prime(-1)=(-1)(3 + 4)=-7<0 \). For \( 0\frac{4}{3} \), say \( x = 2 \), \( f^\prime(2)=2\times(-6 + 4)=-4<0 \). So at \( x = 0 \), it's a local minimum? Wait, no, the sign of derivative: left of 0 (x=-1): f’(x)<0 (function decreasing). Between 0 and 4/3 (x = 1): f’(x)>0 (function increasing). So at x = 0, the function has a local minimum (since it was decreasing, then increasing). At x = 4/3, it's a local maximum.

But maybe for sketching, we can use the key points: (0,0) (touch), (2,0) (cross), (1,1), (-1,3), and end - behavior. So the graph starts from the top left (x→-∞, f(x)→+∞), comes down to (0,0) (touch, local min), then goes up to a local max at x = 4/3, then comes down to cross at (2,0) and then goes to -∞ as x→+∞.

Step1: Recall transformation rules for functions

For a function \( y = a(x - h)^n + k \) from the parent function \( y=x^n \):

• \( h \): horizontal shift (right if \( h>0 \), left if \( h < 0 \))
• \( a \): vertical stretch (if \( |a|>1 \)) or compression (if \( 0<|a|<1 \)) and reflection over the x - axis (if \( a < 0 \))
• \( k \): vertical shift (up if \( k>0 \), down if \( k < 0 \))

Step2: Analyze \( g(x)=-3(x - 5)^3 \) and \( f(x)=x^3 \)

• Horizontal shift: The \( (x - 5) \) part means a horizontal shift of 5 units to the right (since \( h = 5>0 \) in the form \( y=a(x - h)^n \))
• Vertical stretch/reflection: The coefficient - 3:
• The absolute value \( |-3| = 3>1 \), so there is a vertical stretch by a factor of 3.
• The negative sign means a reflection over the x - axis.
• There is no vertical shift ( \( k = 0 \) )

Answer:

B. As \( x \to +\infty,f(x) \to -\infty \) and as \( x \to -\infty,f(x) \to -\infty \)

Question 2