1.
Item 1. Write the inequality that represents the shaded region, given the equation of the boundary line: \(y=-\frac{1}{2}x+4\)
Solution:The boundary line is \(y=-\frac{1}{2}x+4\).The graph is shaded below the boundary line, so we use a less than inequality.The boundary line is solid, so the points on the line are included in the solution.Therefore, we use \(\le\), not \(<\).The inequality is:\(y\le -\frac{1}{2}x+4\)Answer: A
2.
Item 3. Which sign analysis statement is true for the following inequality? \(16x^2+8x-3
Solution:Start with the inequality:\(16x^2+8x-3<0\)Factor the quadratic:\(16x^2+8x-3=16x^2+12x-4x-3\)\(=4x(4x+3)-1(4x+3)\)\(=(4x+3)(4x-1)\)Find the zeros:\(4x+3=0\Rightarrow x=-\frac{3}{4}\)\(4x-1=0\Rightarrow x=\frac{1}{4}\)Because the inequality is \(<0\), we need the interval where the product is negative.Test \(x=0\), which lies between \(-\frac{3}{4}\) and \(\frac{1}{4}\):\((4(0)+3)(4(0)-1)=(3)(-1)=-3<0\)So the solution is between the two zeros.Since the inequality is strictly less than zero, the endpoints are not included.\(\left(-\frac{3}{4},\frac{1}{4}\right)\)Answer: B
3.
Item 5. Determine an inequality for the following graph.
Solution:The graph is a downward-opening parabola, so the coefficient of the squared term is negative.From the graph, the zeros are \(-3\) and \(2\).Use the zeros to write the equation in factored form:\(y=-(x+3)(x-2)\)Expand:\(y=-(x^2+x-6)\)\(y=-x^2-x+6\)Convert to vertex form by completing the square:\(y=-(x^2+x)+6\)\(y=-\left(x^2+x+\frac{1}{4}-\frac{1}{4}\right)+6\)\(y=-\left(x+\frac{1}{2}\right)^2+\frac{1}{4}+6\)\(y=-\left(x+\frac{1}{2}\right)^2+6\frac{1}{4}\)The graph is shaded outside/above the parabola, so we use \(>\).The parabola is dotted, so the boundary is not included.Therefore:\(y> -\left(x+\frac{1}{2}\right)^2+6\frac{1}{4}\)Answer: D
4.
Item 4. Solve the inequality by graphing. \(y>2(x-4)^2-1\)
Solution:The boundary equation is:\(y=2(x-4)^2-1\)This is in vertex form:\(y=a(x-h)^2+k\)So the vertex is:\((4,-1)\)Since \(a=2\), the parabola opens upward and has a vertical stretch by a factor of \(2\).Because the inequality is \(y>2(x-4)^2-1\), the boundary line should be dotted, since \(>\) does not include the boundary.Use a test point, such as \((0,0)\):\(0>2(0-4)^2-1\)\(0>2(-4)^2-1\)\(0>2(16)-1\)\(0>32-1\)\(0>31\), which is false.So \((0,0)\) is not in the solution region.The solution must be the region above/inside the upward-opening parabola, matching Graph A.Answer: A
5.
Item 2. Which system of linear inequalities is shown by the following graph?
Solution:First look at the red line.The red line has a \(y\)-intercept of \(-2\), so its equation has the form \(y=mx-2\).From point A to point B, the rise is \(+4\) and the run is \(+4\).So the slope is:\(m=\frac{4}{4}=1\)Therefore, the red boundary line is \(y=x-2\).The red line is solid and the favourable region is above the line, so the inequality is:\(y\ge x-2\)Now look at the brown line.The brown line has a \(y\)-intercept of \(5\), so its equation has the form \(y=mx+5\).From point C to point D, the rise is \(+3\) and the run is \(+2\).So the slope is:\(m=\frac{3}{2}\)Therefore, the brown boundary line is \(y=\frac{3}{2}x+5\).The brown line is dotted and the favourable region is below the line, so the inequality is:\(y<\frac{3}{2}x+5\)The system is:\(y\ge x-2;\; y<\frac{3}{2}x+5\)Answer: B
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