Question

7 which system of inequalities is graphed below?
a (\begin{cases} y > x^2 - 7x - 8 \\ y < -x + 7 end{cases})
b (\begin{cases} y < x^2 + 7x - 8 \\ y > x + 7 end{cases})
c (\begin{cases} y < x^2 - 7x - 8 \\ y > -x + 7 end{cases})
d (\begin{cases} y > x^2 + 7x - 8 \\ y < x + 7 end{cases})

Explanation:

Step1: Analyze the parabola

The parabola has a vertical axis (since it's a quadratic in \(x\)). The general form of a quadratic is \(y = ax^2+bx + c\). Let's find its roots. From the graph, the parabola crosses the \(x\)-axis at \(x = - 1\) and \(x = 8\) (wait, no, looking at the grid, when \(x = 8\), \(y=0\), and when \(x=-1\), \(y = 0\)? Wait, no, let's check the vertex. Wait, the parabola opens upwards (since the dotted part is inside? Wait, no, the solid/dotted lines. Wait, the parabola: let's find its equation. The roots are \(x=-1\) and \(x = 8\)? Wait, no, when \(x = 8\), \(y = 0\), and when \(x=-1\), \(y=0\)? Wait, the quadratic can be written as \(y=(x + 1)(x - 8)=x^2-7x - 8\). Yes, because \((x + 1)(x - 8)=x^2-8x+x - 8=x^2-7x - 8\). Now, the region for the parabola: the shaded (or the region defined by the inequality) is above or below? The parabola is a dotted line (so the inequality is strict) and the region above or below? Let's take a test point, say \((0,0)\). Plug into \(y=x^2-7x - 8\): \(0^2-7(0)-8=-8\). So at \(x = 0\), the parabola has \(y=-8\). The region for the parabola: if we look at the graph, the area above the parabola? Wait, no, let's check the options. Option A has \(y>x^2 - 7x - 8\), option C has \(y<x^2 - 7x - 8\). Let's take a point in the region of the parabola. Let's take \((0,-10)\). Plug into \(y=x^2-7x - 8\): \(y=-8\) at \(x = 0\). So \(-10<-8\), so if the region is below the parabola? Wait, no, the parabola's inequality: the line is dotted, so the inequality is strict. Wait, the parabola in the graph: the region defined by the parabola's inequality. Let's check the line \(y=-x + 7\). Let's find its slope and intercept. The line \(y=-x + 7\) has a slope of \(-1\) and \(y\)-intercept at \(7\). From the graph, the line crosses the \(y\)-axis at \(y = 7\) (since when \(x = 0\), \(y = 7\)) and has a negative slope. Now, take a test point, say \((0,0)\). Plug into \(y=-x + 7\): \(0<7\), so the region for the line \(y=-x + 7\): the line is dotted (so strict inequality) and the region below or above? Let's see, the line \(y=-x + 7\) at \(x = 0\) is \(7\), and the region in the graph: the area below the line \(y=-x + 7\)? Wait, no, let's check the options. Option A: \(y>x^2-7x - 8\) and \(y<-x + 7\). Let's check the parabola: \(x^2-7x - 8\), at \(x = 0\), \(y=-8\). The region above the parabola would be \(y>-8\) at \(x = 0\), and the line \(y=-x + 7\) at \(x = 0\) is \(7\), so \(y<-x + 7\) at \(x = 0\) is \(y<7\). Let's take the point \((0,0)\). Plug into \(y>x^2-7x - 8\): \(0>-8\), which is true. Plug into \(y<-x + 7\): \(0<7\), which is true. Now check the other options. Option C: \(y<x^2-7x - 8\), at \(x = 0\), \(y< - 8\), but \((0,0)\) has \(y = 0\) which is not less than \(-8\), so C is out. Option B: \(y<x^2 + 7x - 8\), the quadratic \(x^2+7x - 8=(x + 8)(x - 1)\), roots at \(x=-8\) and \(x = 1\), which doesn't match the graph (graph has roots at \(x=-1\) and \(x = 8\)), so B is out. Option D: \(y>x^2 + 7x - 8\), same quadratic as B, roots at \(-8\) and \(1\), not matching, so D is out. So the parabola is \(y=x^2-7x - 8\) and the inequality for the parabola is \(y>x^2-7x - 8\) (since the region above the parabola, as tested with \((0,0)\)), and the line is \(y=-x + 7\) with inequality \(y<-x + 7\) (since \((0,0)\) satisfies \(y<-x + 7\)). So the system is \(\begin{cases}y>x^2-7x - 8\\y<-x + 7\end{cases}\), which is option A.

Step2: Analyze the line

The line is \(y=-x + 7\) (slope \(-1\), \(y\)-intercept \(7\)). The inequality for the line: the line is dotted (so strict inequality) and the region below the line (since \((0,0)\) is below \(y=-x + 7\) as \(0<7\)). The parabola's inequality: \(y>x^2-7x - 8\) as tested with \((0,0)\) (since \(0>-8\)).

Answer:

A. \(

$$\begin{cases}y > x^2 - 7x - 8\\y < -x + 7\end{cases}$$

\)