Question

answer attempt 1 out of 3
$f(x) = \

$$\begin{cases} \\square \\text{ for } \\square \\\\ \\square \\text{ for } \\square \\end{cases}$$

$
(there are also buttons for <, >, ≤, ≥, =, ≠, add rule, remove rule, and a submit answer button.)

Explanation:

Step1: Analyze left piece

The left line passes through \(x = -1\) (open circle, \(y = 6\)) and intersects the x - axis at \(x=-2\) (since when \(y = 0\), from the slope calculation: slope \(m=\frac{6 - 0}{-1-(-2)}=\frac{6}{1}=6\)? Wait, no, let's take two points. Let's see, when \(x=-1\), \(y = 6\) (open circle), and when \(x=-2\), \(y = 0\) (intersection with x - axis). Wait, actually, the left line: let's find the equation. The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take two points: when \(x=-1\), \(y = 6\) (open circle), and when \(x\) approaches \(-\infty\), the line goes down. Wait, another way: the left line has a steep slope. Wait, actually, let's look at the x - values. The left piece is for \(x < -1\)? Wait, no, the open circle is at \(x=-1\) (y = 6) and the other open circle is at \(x = 2\) (y = 7). Wait, no, the graph has two pieces: one from the left (with open circle at \(x=-1\), \(y = 6\)) and one from \(x = 2\) (open circle at \(x = 2\), \(y = 7\)) to \(x = 9\) (where it hits the x - axis). Wait, first, find the equation of the left line. Let's take two points on the left line: let's say when \(x=-1\), \(y = 6\) (open circle, so not included) and when \(x=-2\), \(y = 0\) (intersection with x - axis). Wait, the slope \(m=\frac{6-0}{-1 - (-2)}=\frac{6}{1}=6\)? Wait, no, if we take a point further left, say when \(x=-3\), let's see the y - value. Wait, the left line is a straight line with slope, let's calculate correctly. Let's take two points: \((-2,0)\) and \((-1,6)\) (but \((-1,6)\) is an open circle). So the equation of the left line: using point - slope form \(y - y_1=m(x - x_1)\). \(m=\frac{6 - 0}{-1-(-2)} = 6\). So \(y-0=6(x + 2)\), so \(y = 6x+12\). This line is for \(x < -1\) (since the open circle is at \(x=-1\), so \(x\) is less than \(-1\)).

Step2: Analyze right piece

The right piece: open circle at \(x = 2\) (y = 7) and intersects the x - axis at \(x = 9\) (y = 0). Let's find the slope: \(m=\frac{0 - 7}{9 - 2}=\frac{-7}{7}=-1\). Using point - slope form with point \((2,7)\) (open circle, so \(x>2\)? Wait, no, the open circle is at \(x = 2\), \(y = 7\), and the line goes down to \((9,0)\). So the equation: \(y - 7=-1(x - 2)\), so \(y=-x + 2+7=-x + 9\). This line is for \(2 < x\leq9\)? Wait, no, the open circle is at \(x = 2\), so \(x>2\) and it ends at \(x = 9\) (where \(y = 0\)). Wait, but let's check the domain. The left piece: open circle at \(x=-1\), so \(x < -1\), equation \(y = 6x + 12\). The right piece: open circle at \(x = 2\), so \(x>2\), equation \(y=-x + 9\) (since when \(x = 9\), \(y=0\), and when \(x = 2\), \(y=-2 + 9=7\), which matches the open circle at \((2,7)\)).

Wait, maybe I made a mistake in the left line. Let's re - check. The left line: when \(x=-1\), \(y = 6\) (open circle), and when \(x\) is less than \(-1\), the line goes down. Wait, maybe the slope is steeper. Wait, looking at the graph, the left line has a slope of, let's see, from \(x=-1\) (y = 6) to \(x=-2\) (y = 0), the change in y is - 6, change in x is - 1, so slope is \(\frac{-6}{-1}=6\). So the equation is \(y=6(x + 2)\) (since it passes through \((-2,0)\)), so \(y = 6x+12\) for \(x < -1\).

The right line: from \(x = 2\) (y = 7, open circle) to \(x = 9\) (y = 0). Slope is \(\frac{0 - 7}{9 - 2}=-1\), so equation is \(y-7=-1(x - 2)\), so \(y=-x + 9\) for \(x>2\).

Answer:

\(f(x)=

$$\begin{cases}6x + 12& \text{for }x < - 1\\-x + 9& \text{for }x>2\end{cases}$$

\)