Question

use a graph to solve the inequality.\\(x^2 - x - 12 > 0\\)\\(\dots\\)to solve the inequality, identify the values of \\(x\\) that make the value of the expression \\(x^2 - x - 12\\) greater than 0.\\(\
ewline\\)graph the equation \\(y = x^2 - x - 12\\) by plotting points. complete the following table.\\(\
ewline\\)\\(\

$$\begin{array}{|c|c|c|c|c|c|} \\hline x & -2 & -1 & 0 & 1 & 2 \\\\ \\hline x^2 - x - 12 & -6 & -10 & -12 & -12 & -10 \\\\ \\hline \\end{array}$$

\\)\\(\
ewline\\)use the points to sketch the graph. choose the correct graph on the right.\\(\
ewline\\)to solve the inequality, look for points on the graph where the value of the function is \\(\boxed{}\\)

Explanation:

Step1: Analyze the function

The function is \( y = x^2 - x - 12 \), which is a quadratic function. The coefficient of \( x^2 \) is positive, so the parabola opens upwards.

Step2: Plot the points

From the table, we have the points \((-2, -6)\), \((-1, -10)\), \((0, -12)\), \((1, -12)\), \((2, -10)\). Let's check the y - intercept: when \( x = 0 \), \( y=- 12 \), so the graph passes through \((0, - 12)\). When \( x = 1 \) and \( x = 0 \), \( y=-12\); when \( x=-1\) and \( x = 2 \), \( y = - 10 \); when \( x=-2\), \( y=-6\).

Step3: Match with the graphs

• For option A: The parabola opens upwards? Wait, no, looking at the shape, the one that has the vertex and passes through the points we calculated. Wait, the correct graph should have a minimum point (since the parabola opens upwards) and pass through the points. Let's check the vertex of \( y=x^2 - x - 12\). The x - coordinate of the vertex is \( x=-\frac{b}{2a}=-\frac{-1}{2\times1}=\frac{1}{2}\), and \( y = (\frac{1}{2})^2-\frac{1}{2}-12=\frac{1}{4}-\frac{1}{2}-12=\frac{1 - 2}{4}-12=-\frac{1}{4}-12=-\frac{49}{4}=- 12.25\). Now, looking at the graphs, the graph in option C? Wait, no, wait the original problem's graph options: Let's re - evaluate. The function \( y=x^2 - x - 12=(x - 4)(x + 3)\), so the roots are at \( x = 4 \) and \( x=-3 \). But from our table, we have points for \( x=-2,-1,0,1,2\). The graph that opens upwards and passes through \((0, - 12)\), \((1, - 12)\), etc. The correct graph should be the one with the upward - opening parabola passing through these points. Wait, among the given graphs, the graph in option C (the one in the middle - top) seems to match the points. Wait, no, let's check the direction. Since the coefficient of \( x^2 \) is positive, the parabola opens upwards. Now, looking at the points, when \( x\) increases from - 2 to 0, \( y\) decreases from - 6 to - 12, and when \( x\) increases from 0 to 2, \( y\) increases from - 12 to - 10. So the vertex is between \( x = 0 \) and \( x = 1 \) (at \( x=\frac{1}{2}\)). The graph that has this behavior is the one in option C? Wait, no, the user's graph options: Let's see, the correct graph should be the one with the upward - opening parabola. Wait, the graph in option C (the third graph in the first row? Wait, the original image: The correct graph is the one that has the points plotted correctly. After plotting the points \((-2,-6)\), \((-1,-10)\), \((0,-12)\), \((1,-12)\), \((2,-10)\), the graph that matches is the one in option C (the graph labeled C).

Answer:

C (the graph labeled C)