Question

week 15 - hw 2
create a table of points, graph and find the rule for each problem.
1.
(0, 4)
(-3, -4)
table:
rule:
2.
rule: y = - 3
table:
3.
rule: x = 3
table:
4.
table:
rule:

Explanation:

Step1: Analyze Problem 1 (Find Table and Rule)

We have two points: \((-3, -4)\) and \((0, 4)\). First, find the slope \(m\) using \(m=\frac{y_2 - y_1}{x_2 - x_1}\). So \(m=\frac{4 - (-4)}{0 - (-3)}=\frac{8}{3}\)? Wait, no, wait the graph: from \((-3, -4)\) to \((0, 4)\), the rise is \(4 - (-4)=8\), run is \(0 - (-3)=3\)? Wait, no, maybe I misread. Wait, the graph: the line goes from \((-3, -4)\) to \((0, 4)\). Wait, let's check the grid. Wait, maybe the horizontal distance (run) is 3 units (from \(x=-3\) to \(x=0\)), vertical distance (rise) is \(4 - (-4)=8\)? No, that can't be. Wait, maybe the points are \((-3, -4)\) and \((0, 4)\), so slope \(m=\frac{4 - (-4)}{0 - (-3)}=\frac{8}{3}\)? Wait, no, maybe the graph is steeper. Wait, alternatively, maybe the points are \((-3, -4)\) and \((0, 4)\), so let's make a table. Let's pick \(x=-3\), \(y=-4\); \(x=0\), \(y=4\); let's find another point. Let \(x=3\), then using slope \(m=\frac{4 - (-4)}{0 - (-3)}=\frac{8}{3}\), so \(y - 4=\frac{8}{3}(x - 0)\), so \(y=\frac{8}{3}x + 4\). Wait, but maybe I made a mistake. Wait, the graph: from \((-3, -4)\) to \((0, 4)\), the vertical change is 8, horizontal change is 3? Wait, no, maybe the grid is 1 unit per square. So from \((-3, -4)\) to \((0, 4)\): \(x\) increases by 3, \(y\) increases by 8? That seems odd. Wait, maybe the points are \((-3, -4)\) and \((0, 4)\), so let's create the table.

Table for Problem 1:

\(x_1\)	\(y_1\)
\(0\)	\(4\)
\(3\)	\(12\)	(using \(y=\frac{8}{3}x + 4\), when \(x=3\), \(y=\frac{8}{3}(3)+4=8 + 4=12\))

Now, find the rule (equation). Using slope-intercept form \(y = mx + b\), \(b\) is the y-intercept, which is 4 (from \((0, 4)\)). Slope \(m=\frac{4 - (-4)}{0 - (-3)}=\frac{8}{3}\)? Wait, no, wait the vertical change: from \((-3, -4)\) to \((0, 4)\), \(y\) goes from -4 to 4, so that's +8, \(x\) from -3 to 0, +3. So slope \(m=\frac{8}{3}\). So the equation is \(y=\frac{8}{3}x + 4\). Wait, but maybe I misread the points. Alternatively, maybe the points are \((-3, -4)\) and \((0, 4)\), so the table is:

\(x=-3\), \(y=-4\); \(x=0\), \(y=4\); \(x=3\), \(y=12\) (as above).

Step2: Problem 2 (Complete Table for \(y = -3\))

The rule is \(y=-3\), so for any \(x\), \(y=-3\). So table:

\(x_1\)	\(y_1\)
\(0\)	\(-3\)
\(2\)	\(-3\)

Step3: Problem 3 (Complete Table for \(x = 3\))

The rule is \(x=3\), so \(x\) is always 3, \(y\) can be any value. Table:

\(x_1\)	\(y_1\)
\(3\)	\(0\)
\(3\)	\(2\)

Step4: Problem 4 (Find Table and Rule)

Points: \((0, 3)\) and \((2, 0)\). Slope \(m=\frac{0 - 3}{2 - 0}=\frac{-3}{2}=-\frac{3}{2}\). Y-intercept \(b=3\) (from \((0, 3)\)). So equation \(y=-\frac{3}{2}x + 3\). Table:

\(x_1\)	\(y_1\)
\(2\)	\(0\)
\(4\)	\(-3\)	(using \(y=-\frac{3}{2}(4)+3=-6 + 3=-3\))

Answer:

Problem 1:

Table:

\(x_1\)	\(y_1\)
\(0\)	\(4\)
\(3\)	\(12\)

Rule: \(y=\frac{8}{3}x + 4\) (or corrected if slope is miscalculated, but based on points \((-3, -4)\) and \((0, 4)\))

Problem 2:

Table (example):

\(x_1\)	\(y_1\)
\(0\)	\(-3\)
\(2\)	\(-3\)

Rule: \(y = -3\) (given)

Problem 3:

Table (example):

\(x_1\)	\(y_1\)
\(3\)	\(0\)
\(3\)	\(2\)

Rule: \(x = 3\) (given)

Problem 4:

Table:

\(x_1\)	\(y_1\)
\(2\)	\(0\)
\(4\)	\(-3\)

Rule: \(y=-\frac{3}{2}x + 3\)