Question

3.7 hw day 2 - solve each system by graphing. please add arrows to all graphs first. quadratic - quadratic 1. \\( \

$$\begin{cases} y \\leq -2x^2 + x + 5 \\\\ y > (x + 1)^2 - 8 \\end{cases}$$

\\) linear - absolute value 2. \\( \

$$\begin{cases} y < -2|x - 1| + 7 \\\\ y \\geq 3x - 5 \\end{cases}$$

\\)

Explanation:

Step1: Analyze the first system (Quadratic - Quadratic)

The first system is \(

$$\begin{cases}y\leq - 2x^{2}+x + 5\\y>(x + 1)^{2}-8\end{cases}$$

\)

• For the parabola \(y=-2x^{2}+x + 5\), since the coefficient of \(x^{2}\) is \(-2<0\), it opens downwards. The inequality \(y\leq - 2x^{2}+x + 5\) represents the region below or on this parabola.
• For the parabola \(y=(x + 1)^{2}-8=x^{2}+2x+1 - 8=x^{2}+2x - 7\), the coefficient of \(x^{2}\) is \(1>0\), so it opens upwards. The inequality \(y>(x + 1)^{2}-8\) represents the region above this parabola (not including the parabola itself).

To graph these, first, find the vertex of \(y=(x + 1)^{2}-8\), the vertex is \((-1,-8)\). For \(y=-2x^{2}+x + 5\), the \(x\)-coordinate of the vertex is \(x =-\frac{b}{2a}=-\frac{1}{2\times(-2)}=\frac{1}{4}\), and the \(y\)-coordinate is \(y=-2\times(\frac{1}{4})^{2}+\frac{1}{4}+5=-2\times\frac{1}{16}+\frac{1}{4}+5=-\frac{1}{8}+\frac{2}{8}+5=\frac{1}{8}+5=\frac{41}{8}\).
Then, find the intersection points of the two parabolas by solving \(-2x^{2}+x + 5=(x + 1)^{2}-8\)

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LATEXBLOCK1
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\(x_1=\frac{-1+\sqrt{145}}{6}\approx\frac{-1 + 12.04}{6}\approx1.84\), \(x_2=\frac{-1-\sqrt{145}}{6}\approx\frac{-1-12.04}{6}\approx - 2.17\)
The solution region is the area that is below the downward - opening parabola and above the upward - opening parabola.

Step2: Analyze the second system (Linear - Absolute Value)

The second system is \(

$$\begin{cases}y<-2|x - 1|+7\\y\geq3x - 5\end{cases}$$

\)

• For the absolute - value function \(y=-2|x - 1|+7\), we consider two cases:
• When \(x\geq1\), \(y=-2(x - 1)+7=-2x + 2 + 7=-2x+9\)
• When \(x<1\), \(y=-2(1 - x)+7=-2 + 2x+7=2x + 5\)

The vertex of the absolute - value graph is at \((1,7)\) (since for \(y = a|x - h|+k\), the vertex is \((h,k)\)). The inequality \(y<-2|x - 1|+7\) represents the region below this absolute - value graph (not including the graph itself).

• For the line \(y = 3x-5\), the slope \(m = 3\) and the \(y\)-intercept \(b=-5\). The inequality \(y\geq3x - 5\) represents the region above or on this line.

To find the intersection point of \(y=-2|x - 1|+7\) and \(y = 3x-5\), we consider two cases:

• Case 1: \(x\geq1\), solve \(-2x + 9=3x-5\), \(5x=14\), \(x=\frac{14}{5}=2.8\), then \(y=3\times\frac{14}{5}-5=\frac{42}{5}-\frac{25}{5}=\frac{17}{5}=3.4\)
• Case 2: \(x<1\), solve \(2x + 5=3x-5\), \(x = 10\) (but \(x = 10\) does not satisfy \(x<1\), so we discard this solution)

So the intersection point is \((\frac{14}{5},\frac{17}{5})\)
The solution region is the area that is below the absolute - value graph and above the line \(y = 3x-5\)

(Note: Since the problem says "PLEASE ADD ARROWS TO ALL GRAPHS FIRST", we need to add arrows to the ends of the graphs to indicate the direction of the lines/parabolas. For the parabolas, the arrows go towards the direction of the opening (downwards for \(y=-2x^{2}+x + 5\) and upwards for \(y=(x + 1)^{2}-8\)). For the absolute - value graph, the two rays (for \(x\geq1\) and \(x<1\)) have arrows at their ends, and for the line \(y = 3x-5\), the arrow goes in the direction of the slope (upwards to the right and downwards to the left). After adding arrows, we can identify the solution regions as described above.)

Answer:

For the first system \(

$$\begin{cases}y\leq - 2x^{2}+x + 5\\y>(x + 1)^{2}-8\end{cases}$$

\), the solution region is the area between the two parabolas (below the downward - opening parabola and above the upward - opening parabola). For the second system \(

$$\begin{cases}y<-2|x - 1|+7\\y\geq3x - 5\end{cases}$$

\), the solution region is the area below the absolute - value graph and above the line \(y = 3x-5\). (The final answer is the graphical solution regions as described, and the process of graphing with arrows is as above.)