Question

algebra 1
unit 3b assignment #2
linear inequalities
name
unit 1/unit 2 review

1. \\( \frac { 3 x + 2 } { 8 } = \frac { 2 x - 4 } { 6 } \\)
2. given \\( h ( x ) = - 3 x + 2 \\) find \\( h ( - 2 ) \\)
3. graph the following:

a. \\( y \leq - x - 2 \\)
b. \\( y > \frac { 3 } { 2 } x - 2 \\)
c. \\( x > 2 \\)

1. use the graph of \\( y = 3 x + 4 \\) for the following parts.

a. shade the graph based on if you were given the
equation \\( y \leq 3 x - 4 \\).
b. circle any of the following points that are solutions
to the inequality.
\\( ( 0,0 ) \\)
\\( ( 4,2 ) \\)
\\( ( - 3,4 ) \\)
\\( ( 3,0 ) \\)
c. is \\( ( 0, - 4 ) \\) a solution to \\( y < 3 x - 4 \\)?

Explanation:

Step1: Substitute \(x = - 2\) into \(h(x)=-3x + 2\)

Substitute \(x=-2\) into the function \(h(x)\). According to the function formula \(h(x)=-3x + 2\), when \(x=-2\), we have \(h(-2)=-3\times(-2)+2\).

Step2: Calculate the value of \(h(-2)\)

First, calculate \(-3\times(-2)=6\). Then \(h(-2)=6 + 2\).

Answer:

\(h(-2)=8\)