1.
Item 1. Write the inequality that represents the shaded region.
Solution:Use points (-4,5) and (4,-1). Slope m=(-1-5)/(4-(-4))=-6/8=-3/4. Equation: y=-3/4x+2. Region is below a solid line, so use ≤.Answer: A
2.
Item 3. Determine which quadratic inequality statement is correct.
Solution:Zeros are x=-1 and x=2. Therefore equation is (x+1)(x-2)=0 ⇒ x²-x-2=0. Solution region is outside the zeros and includes endpoints, so ≥0.Answer: C
3.
Item 4. Solve \(3x^2-11x-14>0\).
Solution:Use quadratic formula on 3x²-11x-14=0. Roots: x=(11±17)/6 ⇒ x=-1 and x=14/3. Since parabola opens upward and inequality is >0, solution is outside the roots.Answer: A
4.
Item 5. Determine an inequality for the graph.
Solution:Vertex is (3,1) and a=2, so equation is y=2(x-3)^2+1. Region is below the parabola and boundary is dotted, so use <.Answer: A
5.
Item 2. Which system of linear inequalities is shown by the graph?
Solution:Pink line: y-intercept 1 and slope 2, dotted with shading below ⇒ y-3/2x-1.Answer: B
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