Question

lesson 3.3 checkpoint
complete the previous problems, check your solut
complete the lesson reflection above by circling y
evaluate the function when x = -2, 0, and 5.

1. v(x) = 12 - 2x - 5

Explanation:

Step1: Simplify the function

First, simplify \( v(x)=12 - 2x - 5 \). Combine like terms: \( 12-5 = 7 \), so \( v(x)=7 - 2x \).

Step2: Evaluate at \( x = -2 \)

Substitute \( x=-2 \) into \( v(x) \): \( v(-2)=7-2(-2) \). Calculate \( -2(-2)=4 \), then \( 7 + 4 = 11 \).

Step3: Evaluate at \( x = 0 \)

Substitute \( x = 0 \) into \( v(x) \): \( v(0)=7-2(0) \). Since \( 2(0)=0 \), we get \( 7-0 = 7 \).

Step4: Evaluate at \( x = 5 \)

Substitute \( x = 5 \) into \( v(x) \): \( v(5)=7-2(5) \). Calculate \( 2(5)=10 \), then \( 7-10=-3 \).

Answer:

When \( x = -2 \), \( v(-2)=11 \); when \( x = 0 \), \( v(0)=7 \); when \( x = 5 \), \( v(5)=-3 \)