Question

options: {y = x² + x - 8, y = -2x - 4}, {y = x² - x - 8, y = 2x - 4}, {y = -x² - x + 8, y = 2x + 4}, {y = -x² - x + 8, y = -2x - 4} (with graphs of parabolas and lines)

Explanation:

Step1: Analyze the parabola's direction

For a quadratic function \( y = ax^2+bx + c \), if \( a>0 \), the parabola opens upward; if \( a < 0 \), it opens downward. Looking at the graphs, the parabola opens downward, so \( a=- 1 \) (since the coefficient of \( x^2 \) should be negative). So we can eliminate options with \( a = 1 \) (first two options: \( y=x^{2}-x - 8 \) and \( y=x^{2}+x - 8 \) have \( a = 1 \), open upward, which doesn't match the downward - opening parabola in the graphs).

Step2: Analyze the linear function's slope and y - intercept

The linear function is in the form \( y=mx + b \). Let's check the slope and y - intercept. For the linear function, we can also check the intersection with the y - axis. The linear function in the correct option should have a positive slope? Wait, no, let's check the options. The remaining options are \( \{y=-x^{2}-x + 8,y = 2x+4\} \) and \( \{y=-x^{2}-x + 8,y=-2x - 4\} \) (wait, no, original options: third option is \( \{y=-x^{2}-x + 8,y=-2x - 4\} \), fourth is \( \{y=-x^{2}-x + 8,y = 2x+4\} \)). Wait, let's check the y - intercept of the linear function. If we look at the graphs, the linear function: let's take the fourth option \( y = 2x+4 \), y - intercept at \( (0,4) \), and the third option \( y=-2x - 4 \) has y - intercept at \( (0,-4) \). Also, the slope: if the linear function is increasing (positive slope) or decreasing (negative slope). From the graph, the linear function seems to have a positive slope? Wait, no, maybe better to check the system of equations. Wait, the parabola is \( y=-x^{2}-x + 8 \) (since \( a=-1 \), opens downward). Now, for the linear function, let's check the intersection points or the slope. Alternatively, let's check the y - intercept of the parabola. For \( y=-x^{2}-x + 8 \), when \( x = 0 \), \( y=8 \), which matches the y - intercept of the parabola (intersects the y - axis at \( y = 8 \)). Now, for the linear function: in the option \( \{y=-x^{2}-x + 8,y = 2x+4\} \), when \( x = 0 \), \( y = 4 \), and in \( \{y=-x^{2}-x + 8,y=-2x - 4\} \), when \( x = 0 \), \( y=-4 \). Looking at the graph, the linear function intersects the y - axis at a positive value? Wait, no, maybe I made a mistake. Wait, let's re - examine. Wait, the correct option should be \( \{y=-x^{2}-x + 8,y = 2x+4\} \)? Wait, no, wait the first two options are eliminated because their parabolas open upward. Now, between the last two: \( \{y=-x^{2}-x + 8,y = 2x+4\} \) and \( \{y=-x^{2}-x + 8,y=-2x - 4\} \). Let's check the slope of the linear function. If the linear function is \( y = 2x+4 \), slope \( m = 2>0 \), line is increasing. If it's \( y=-2x - 4 \), slope \( m=-2<0 \), line is decreasing. From the graph, the linear function seems to be increasing (going from bottom - left to top - right), so slope is positive. So the linear function should be \( y = 2x+4 \), and the parabola is \( y=-x^{2}-x + 8 \). So the correct option is \( \{y=-x^{2}-x + 8,y = 2x+4\} \) (the fourth option in the list: the one with \( y=-x^{2}-x + 8 \) and \( y = 2x+4 \)).

Answer:

\(\{y=-x^{2}-x + 8,y = 2x+4\}\) (the fourth option in the given set of options, i.e., the option with \( y=-x^{2}-x + 8 \) and \( y = 2x+4\))