Algebra 1 · Unit 2
Graph a linear function by hand from slope and intercepts, with a table, and in a real-life window.
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From slope-intercept
Plot (0, b). From there, use rise over run. Draw the line through the points.
From intercepts
Plot the x-intercept and the y-intercept, then connect them. Works well for Ax + By = C.
A table is a technology graph
A calculator table lists x and y. Plot those points. Equal x-steps with a constant y-change confirm a line.
Real-life graphs
Label the axes with units. Often only the first quadrant is used (time and amount cannot be negative).
Plot from m and b, or from a table.
Y-intercept (0, −1)
Slope 2 = 21, so rise 2, run 1.
Another point: (1, 1)
Answer: through (0, −1) and (1, 1)
Start at (0, 2).
Rise 1, run 3, to (3, 3).
Answer: through (0, 2) and (3, 3)
(4, 0) and (0, 4)
Answer: through (4, 0) and (0, 4)
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| y | 5 | 4 | 3 | 2 |
Points (0, 5), (1, 4), (2, 3), (3, 2)
Answer: line through (0, 5) with slope −1
They meet at (−2, 3).
Answer: y = 3 and x = −2
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| A | 40 | 50 | 60 | 70 | 80 |
Only t ≥ 0 is used in the story. Start at (0, 40), slope 10.
Answer: ray from (0, 40) with slope 10
Negative slopes, standard form, and a story in the first quadrant.
Y-intercept (0, 6)
Zero: 0 = −2x + 6, x = 3, so (3, 0)
Extra point: x = 1, y = 4
Answer: through (0, 6) and (3, 0)
y = 0: 2x = 6, (3, 0)
x = 0: −3y = 6, y = −2, (0, −2)
Answer: through (3, 0) and (0, −2)
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| y | −1 | 0 | 1 | 2 |
Points (0, −1), (2, 0), (4, 1), (6, 2)
Equal x-steps of 2, y-steps of 1, so slope 12.
Answer: through (0, −1) with slope 12
At h = 0, C = 8. At h = 5, C = 28.
Draw the segment from (0, 8) to (5, 28). Do not continue into negative hours.
Answer: segment from (0, 8) to (5, 28)
(0, 1) and, run 2 down 1, (2, 0)
Answer: (0, 1) and (2, 0)
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| 3x − 2 | −2 | 1 | 4 | 7 |
First differences 3, 3, 3 are constant, so the points lie on a line with slope 3.
Answer: constant first differences of 3
Scale, restricted domain, and choosing a window.
C(0) = 12, C(8) = 16
x from about −1 to 9, y from about 10 to 18, so the segment is visible (not a tiny sliver at the top of a −10 to 10 window).
Plot (0, 12) and (8, 16). Slope 12.
Answer: window around 0 to 8 and 12 to 16
5y = −2x
y = −25x
Through (0, 0). Run 5, down 2, to (5, −2).
Answer: through (0, 0) and (5, −2)
3t + 20 = 50
3t = 30
t = 10
Graph the segment from (0, 20) to (10, 50). After t = 10 the model stops.
Answer: segment to (10, 50)
Y-intercept (0, −8) is below y = −5.
X-intercept (2, 0) is on the window.
A table with x = 0 shows y = −8, so you know to zoom out.
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| y | −8 | −4 | 0 | 4 |
Answer: may miss (0, −8); table shows it
x = −1x + 4
2x = 4
x = 2, y = 2
Answer: (2, 2)
Y-intercept (0, −3) is friendly.
From there, run 7, rise 2, to (7, −1).
Starting at a non-integer point would make counting squares harder. A table with x = 0 and x = 7 does the same work.
Answer: (0, −3) and (7, −1)
On that window, y is always about 19 to 21, so the line is above the view. You see a blank screen.
Change Ymin to about 15 and Ymax to about 25, or zoom out until (0, 20) is visible.
A table immediately shows y ≈ 20, which tells you the window is wrong, not the equation.
Answer: blank window; raise the y-range to include 20
The story is a segment from (0, 15) to (6, 27).
The infinite line also has t = −2, C = 11 (negative time) and t = 10, C = 35 (after the 6-hour window).
Those points solve the equation but are not in the story’s domain.
Answer: segment vs whole line
The table 2, 4, 6 matches y = 2x + 2, a line.
(0, 2), (1, 4), (2, 5) has slopes 2 then 1, so not linear.
The third plotted point was likely a counting error (up 1 instead of up 2).
Answer: the table is linear; (2, 5) is off the line
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