Question

7
f(x) = (x + 6)(x + 5)(x - 4)
the function f is given. which table of values represents y = f(x) - 3?

option a:

x	y
-6	-9
-5	-8
4	1

option b:

x	y
-6	-3
-5	-3
4	-3

option c:

x	y
-6	-3
-5	-2
4	7

option d:

x	y
-6	3
-5	3
4	3

Explanation:

Step1: Find f(-6)

Substitute \( x = -6 \) into \( f(x)=(x + 6)(x + 5)(x - 4) \).
\( f(-6)=(-6 + 6)(-6 + 5)(-6 - 4)=0\times(-1)\times(-10)=0 \).
Then \( y = f(-6)-3 = 0 - 3=-3 \).

Step2: Find f(-5)

Substitute \( x = -5 \) into \( f(x) \).
\( f(-5)=(-5 + 6)(-5 + 5)(-5 - 4)=1\times0\times(-9)=0 \).
Then \( y = f(-5)-3 = 0 - 3=-3 \)? Wait, no—wait, option A: when \( x=-5 \), \( y=-8 \)? Wait, no, let's recalculate. Wait, no, the function is \( y = f(x)-3 \). Wait, maybe I made a mistake. Wait, no, let's check option A again. Wait, no, let's recalculate \( f(-5) \): \( (-5 + 6)=1 \), \( (-5 + 5)=0 \), \( (-5 - 4)=-9 \), so \( f(-5)=1\times0\times(-9)=0 \), so \( y = 0 - 3=-3 \). But option A has \( x=-5 \), \( y=-8 \). Wait, maybe I misread the function. Wait, the function is \( f(x)=(x + 6)(x + 5)(x - 4) \). Wait, no, maybe the original function is \( f(x)=(x + 6)(x + 5)(x - 4) \), so for \( x = -6 \), \( f(-6)=0 \), so \( y = 0 - 3=-3 \). For \( x=-5 \), \( f(-5)=0 \), so \( y = 0 - 3=-3 \). For \( x = 4 \), \( f(4)=(4 + 6)(4 + 5)(4 - 4)=10\times9\times0=0 \), so \( y = 0 - 3=-3 \)? But option B has \( x=4 \), \( y=-3 \). Wait, but let's check option A: \( x=-6 \), \( y=-3 \)? No, option A has \( x=-6 \), \( y=-9 \). Wait, maybe I messed up the function. Wait, the problem is \( y = f(x)-3 \), so \( f(x) \) values:

Wait, no, let's re-express. Let's compute \( f(x) \) at \( x=-6 \), \( x=-5 \), \( x=4 \):

• \( x=-6 \): \( f(-6)=(-6 + 6)(-6 + 5)(-6 - 4)=0\times(-1)\times(-10)=0 \). So \( y = 0 - 3=-3 \).
• \( x=-5 \): \( f(-5)=(-5 + 6)(-5 + 5)(-5 - 4)=1\times0\times(-9)=0 \). So \( y = 0 - 3=-3 \).
• \( x=4 \): \( f(4)=(4 + 6)(4 + 5)(4 - 4)=10\times9\times0=0 \). So \( y = 0 - 3=-3 \).

Looking at the options, option B has \( x=-6 \), \( y=-3 \); \( x=-5 \), \( y=-3 \); \( x=4 \), \( y=-3 \). Wait, but that can't be, because when \( x=4 \), \( f(4)=0 \), so \( y=0 - 3=-3 \). So option B matches? Wait, but let's check the options again. Wait, the options are:

A: \( x=-6, y=-9 \); \( x=-5, y=-8 \); \( x=4, y=1 \)

B: \( x=-6, y=-3 \); \( x=-5, y=-3 \); \( x=4, y=-3 \)

C: \( x=-6, y=-3 \); \( x=-5, y=-2 \); \( x=4, y=7 \)

D: \( x=-6, y=3 \); \( x=-5, y=3 \); \( x=4, y=3 \)

Wait, but when \( x=4 \), \( f(4)=(4 + 6)(4 + 5)(4 - 4)=10\times9\times0=0 \), so \( y = 0 - 3=-3 \), which is option B. But wait, maybe I made a mistake in \( f(4) \). Wait, \( (4 - 4)=0 \), so yes, \( f(4)=0 \), so \( y=0 - 3=-3 \). And \( f(-6)=0 \), \( y=-3 \); \( f(-5)=0 \), \( y=-3 \). So option B has all three points: \( (-6, -3) \), \( (-5, -3) \), \( (4, -3) \), which matches the calculations.

Wait, but earlier I thought option A, but no—wait, no, my initial calculation for \( x=-5 \) was wrong? Wait, no, \( f(-5)=0 \), so \( y = 0 - 3=-3 \), which is option B. So the correct table is option B.

Answer:

B.

x	y
-5	-3
4	-3