Algebra 1
Linear Functions & Their Graphs
Fresh, parallel-form problems with full worked solutions — more reps for the skills in this unit, kept separate from the textbook's own problems.
Lesson 5.1: The coordinate plane
Without plotting, name the quadrant (or axis) where (6, -3) lies, and say how the signs tell you.
Worked solutionTry it first, then open.
Read the address x first, then y. Here x = +6 (right) and y = -3 (down). The sign pattern (+, -) is Quadrant IV (bottom-right). Right then down lands you in the lower-right region, so (6, -3) is in Quadrant IV.
Answer: Quadrant IV
Where does (0, -5) live, and why is it NOT in a quadrant?
Worked solutionTry it first, then open.
The x-coordinate is 0, so there is no left/right move at all — you stay on the vertical line that is the y-axis, then go down 5. A point only sits in a quadrant when BOTH coordinates are nonzero; a 0 coordinate pins the point onto an axis. So (0, -5) is on the y-axis, not in any quadrant. (This is the classic axis-point trap.)
Answer: on the y-axis (no quadrant)
I am thinking of a point whose x-coordinate is negative and whose y-coordinate is positive. Which quadrant must it be in, and give one example point that fits.
Worked solutionTry it first, then open.
Negative x means left of the y-axis; positive y means above the x-axis. Left-and-up is the top-left region, Quadrant II, whose sign pattern is (-, +). Any point with a negative first coordinate and a positive second coordinate works, for example (-3, 5).
Answer: Quadrant II; e.g. (-3, 5)
The point (-4, 7) is in Quadrant II. Reflect it across the x-axis (flip it straight up/down to the mirror-image point below). What are the new coordinates, and which quadrant is the reflection in?
Worked solutionTry it first, then open.
Reflecting across the x-axis keeps the left/right position the same and flips the up/down sign: x stays -4, y becomes the opposite of 7, which is -7. So the image is (-4, -7). Its signs are (-, -) — left and down — which is Quadrant III. (Sanity check: a point above the x-axis in Quadrant II reflects to a point below it; left of the y-axis stays left, so II goes to III.)
Answer: (-4, -7), Quadrant III
Lesson 5.2: Graphing linear equations from a table
Build a table for y = 4x - 3. Find y when x = 2.
Worked solutionTry it first, then open.
Feed x = 2 into the machine y = 4x - 3: y = 4(2) - 3 = 8 - 3 = 5. The point is (2, 5).
Answer: 5
Build a table for y = -2x + 5. Find y when x = -1 (watch the negatives).
Worked solutionTry it first, then open.
Substitute x = -1: y = -2(-1) + 5. A negative times a negative is positive, so -2(-1) = +2, then 2 + 5 = 7. The point is (-1, 7). The trap is dropping a sign and getting 3; keep the double negative.
Answer: 7
Without graphing, is the point (3, 7) on the line y = 2x + 1? Show how you know.
Worked solutionTry it first, then open.
A point is on the line exactly when its coordinates make the equation true. Substitute x = 3: 2(3) + 1 = 6 + 1 = 7. The equation predicts y = 7, and the point's y is 7. Since 7 = 7, the point checks — (3, 7) is on the line. (Membership is exact, not 'close.')
Answer: Yes, it is on the line
For y = ½x + 3, find y when x = -4 (a fractional slope with a negative input).
Worked solutionTry it first, then open.
Substitute x = -4: y = ½(-4) + 3. Half of -4 is -2, then -2 + 3 = 1. The point is (-4, 1). Because the slope is ½, even x-values give whole-number y-values, which makes this an easy point to plot.
Answer: 1
Lesson 5.3: Slope
Find the slope of the line through (1, 3) and (5, 11).
Worked solutionTry it first, then open.
Use m = (y₂ - y₁)/(x₂ - x₁) with point 1 = (1, 3) and point 2 = (5, 11), subtracting in the same order top and bottom: m = (11 - 3)/(5 - 1) = 8/4 = 2. Positive slope, so the line goes uphill left-to-right; y rises 2 for every 1 step in x.
Answer: 2
Find the slope of the line through (-2, 4) and (2, -4).
Worked solutionTry it first, then open.
With point 1 = (-2, 4) and point 2 = (2, -4): m = (-4 - 4)/(2 - (-2)) = -8/4 = -2. Keep the subtraction order consistent: the bottom is 2 - (-2) = 4, not 0. The slope is negative, so the line goes downhill; y drops 2 for each 1 step right.
Answer: -2
Find the slope of the line through (2, 1) and (8, 5). Leave it as a reduced fraction.
Worked solutionTry it first, then open.
m = (5 - 1)/(8 - 2) = 4/6. Reduce by dividing top and bottom by 2: 4/6 = 2/3. A slope of 2/3 means y rises 2 for every 3 steps in x — a fairly gentle uphill (rise smaller than run).
Answer: 2/3
On a distance(miles)-vs-time(hours) graph, a car's line passes through (1, 30) and (4, 120). Find the slope WITH its units, and say in words what it means.
Worked solutionTry it first, then open.
Slope is the rate of change, and on this graph the axes are miles (vertical) over hours (horizontal). m = (120 - 30)/(4 - 1) = 90/3 = 30, and reading the units off the axes makes it 30 miles per hour. In words: the distance grows 30 miles for each additional hour — the car is moving at a steady speed of 30 mph. (Constant-rate check: this same rate holds between any two points on the line.)
Answer: 30 miles per hour — the car travels 30 miles each hour (a steady speed of 30 mph)
Lesson 5.4: Slope-intercept form y = mx + b
Identify the slope and the y-intercept of y = -x + 8. Which way does the line tilt?
Worked solutionTry it first, then open.
In y = mx + b, m rides with the x and b stands alone. Here -x means the coefficient is -1, so the slope is m = -1 (the sign travels with the number — the trap is calling it 1). The constant is +8, so the y-intercept is the point (0, 8). Because the slope is negative, the line tilts downhill left-to-right.
Answer: slope -1, y-intercept (0, 8); tilts downhill
For y = 3x - 7, find y when x = 4.
Worked solutionTry it first, then open.
Substitute x = 4: y = 3(4) - 7 = 12 - 7 = 5. So the point (4, 5) is on the line. (Slope 3, intercept (0, -7), if you want to read the form too.)
Answer: 5
For f(x) = ⅔x + 1, find f(6).
Worked solutionTry it first, then open.
f(6) = ⅔(6) + 1. Two-thirds of 6 is 4, then 4 + 1 = 5, so f(6) = 5 and the point (6, 5) is on the line. Choosing a multiple of 3 for x clears the fraction cleanly.
Answer: 5
A pool drains according to d = -15t + 240, where t is minutes and d is gallons left. What do the -15 and the 240 mean about the pool? Is it filling or draining, and how fast? How much water is left after 8 minutes?
Worked solutionTry it first, then open.
Read y = mx + b in context: b is the starting value and m is the rate. The intercept b = 240 means at t = 0 there are 240 gallons — the starting amount. The slope m = -15 means the amount changes by -15 gallons each minute; the negative sign says it is decreasing, so the pool is draining at 15 gallons per minute. After 8 minutes: d = -15(8) + 240 = -120 + 240 = 120 gallons left.
Answer: 240 = starting amount (240 gallons at t = 0); -15 = rate of -15 gallons/min (draining at 15 gal/min); after 8 minutes, 120 gallons left
Lesson 5.5: Writing equations of lines
Find the slope of the line through (2, -1) and (5, 8). (You will use this slope in the next problem to write the line's equation.)
Worked solutionTry it first, then open.
m = (y₂ - y₁)/(x₂ - x₁) = (8 - (-1))/(5 - 2) = 9/3 = 3. Watch the top: 8 - (-1) = 9, not 7. The slope is 3.
Answer: 3
The line through (2, -1) and (5, 8) has slope 3 (from the previous problem). Find its y-intercept value b, then state the slope-intercept equation.
Worked solutionTry it first, then open.
Once the slope is known, the shortcut b = y₁ - m·x₁ uses one point. With (x₁, y₁) = (2, -1) and m = 3: b = -1 - 3(2) = -1 - 6 = -7. So the equation is y = 3x - 7. Check with the other point (5, 8): 3(5) - 7 = 15 - 7 = 8. (Equivalently, point-slope: y - (-1) = 3(x - 2) ⇒ y = 3x - 6 - 1 = 3x - 7.)
Answer: -7
What is the slope of any line perpendicular to y = 4x - 1?
Worked solutionTry it first, then open.
Perpendicular slopes are negative reciprocals: flip the number AND change its sign. The given slope is 4 = 4/1; flip to ¼, then negate to get -¼. Check: 4 · (-¼) = -1, which is the perpendicular test. So the perpendicular slope is -1/4. (The trap is doing only half — using -4 or ¼ alone.)
Answer: -1/4
Classify each pair as parallel, perpendicular, or neither, and explain how you can tell without graphing:
(a) y = -2x + 5 and y = ½x - 3
(b) y = -2x + 5 and y = -2x + 1
Worked solutionTry it first, then open.
Compare slopes only. (a) Slopes -2 and ½: their product is (-2)(½) = -1, the perpendicular test, so the lines are perpendicular (one is the negative reciprocal of the other: flip -2 to -½, negate to +½). (b) Slopes -2 and -2 are equal, but the y-intercepts (5 vs. 1) differ, so the lines never meet — they are parallel. (Equal slope AND equal intercept would make them the same line; equal slope with different intercept is parallel.)
Answer: (a) perpendicular; (b) parallel
Lesson 5.6: x-intercepts, graphing by intercepts & standard form
Find the x-intercept of y = 2x - 8 (set y = 0 and solve for x).
Worked solutionTry it first, then open.
The x-intercept is where the line crosses the x-axis, and every point there has y = 0. Set y = 0: 0 = 2x - 8. Add 8 to both sides: 2x = 8. Divide by 2: x = 4. The x-intercept is the point (4, 0).
Answer: 4
Find the x-intercept of y = ½x - 4 (set y = 0 and solve for x).
Worked solutionTry it first, then open.
Set y = 0: 0 = ½x - 4. Add 4: ½x = 4. Multiply both sides by 2 (undo the ½): x = 8. The x-intercept is (8, 0).
Answer: 8
A line is given in standard form 4x + 3y = 24. Find its x-intercept by setting y = 0 and solving for x.
Worked solutionTry it first, then open.
For the x-intercept, set y = 0 (NOT x = 0 — that would give the y-intercept). Substituting: 4x + 3(0) = 24, so 4x = 24, hence x = 6. The x-intercept is (6, 0). (For contrast, the y-intercept sets x = 0: 3y = 24 ⇒ y = 8, the point (0, 8).)
Answer: 6
Find the x-intercept of y = -⅔x + 4 (set y = 0 and solve for x — a negative fractional slope).
Worked solutionTry it first, then open.
Set y = 0: 0 = -⅔x + 4. Add ⅔x to both sides (or subtract 4 then divide): ⅔x = 4. Multiply both sides by 3/2 (the reciprocal of ⅔): x = 4 · 3/2 = 6. The x-intercept is (6, 0).
Answer: 6
Mixed review
Problems that mix skills from across the unit — good for spacing earlier work back in.
[Slope, 5.3] Find the slope of the line through (-4, 2) and (2, -7). Give a reduced fraction.
Worked solutionTry it first, then open.
m = (y₂ - y₁)/(x₂ - x₁) = (-7 - 2)/(2 - (-4)) = -9/6. Reduce by 3: -9/6 = -3/2. The slope is -3/2 (negative, so downhill; y drops 3 for every 2 steps right).
Answer: -3/2
[Evaluate / point on a line, 5.2 & 5.4] For y = -2x + 9, find y when x = 5. Then state the point (x, y) you found.
Worked solutionTry it first, then open.
Substitute x = 5: y = -2(5) + 9 = -10 + 9 = -1. So y = -1 and the point (5, -1) lies on the line.
Answer: -1
[x-intercept, 5.6] Find the x-intercept of y = 3x - 21.
Worked solutionTry it first, then open.
Set y = 0: 0 = 3x - 21. Add 21: 3x = 21. Divide by 3: x = 7. The x-intercept is (7, 0).
Answer: 7
[Writing equations, 5.5] A line passes through (-2, 5) with slope -4. Find its y-intercept value b (use b = y₁ - m·x₁), then state the slope-intercept equation.
Worked solutionTry it first, then open.
Use b = y₁ - m·x₁ with (x₁, y₁) = (-2, 5) and m = -4: b = 5 - (-4)(-2). First (-4)(-2) = +8, so b = 5 - 8 = -3. The equation is y = -4x - 3. Check at (-2, 5): -4(-2) - 3 = 8 - 3 = 5.
Answer: -3
[Standard form + parallel, 5.6 & 5.5] A line is given by 2x - 3y = 12. (a) Find both intercepts. (b) Solve for y to get slope-intercept form, and state the slope and y-intercept. (c) Is this line parallel to y = ⅔x + 1? Explain.
Worked solutionTry it first, then open.
(a) Zero out the other variable. x-intercept (y = 0): 2x = 12 ⇒ x = 6, point (6, 0). y-intercept (x = 0): -3y = 12 ⇒ y = -4, point (0, -4). (b) Solve for y: 2x - 3y = 12 ⇒ -3y = -2x + 12 ⇒ divide by -3 ⇒ y = ⅔x - 4. So the slope is ⅔ and the y-intercept is (0, -4) — matching part (a). (c) The given line y = ⅔x + 1 also has slope ⅔. Equal slopes with different y-intercepts (-4 vs. 1) means the lines never meet, so yes, they are parallel.
Answer: (a) x-intercept (6, 0), y-intercept (0, -4); (b) y = ⅔x - 4, slope ⅔, y-intercept (0, -4); (c) yes, parallel