Algebra I

Lesson 1: Multi-Step Linear Equations and Inequalities

A linear equation in one variable, like 3x + 7 = 22, asks: what value of x makes this true? Solving means isolating x with a sequence of inverse operations, always keeping the equation balanced. The general strategy: clear parentheses with the distributive property, combine like terms on each side, move variable terms to one side and constants to the other, then divide by the coefficient. Each step is justified by a property of equality — adding, subtracting, multiplying, or dividing both sides by the same (nonzero) number preserves truth.

Some equations need interpretation, not just mechanics. An equation like 2(x + 3) = 2x + 6 simplifies to 6 = 6, a statement always true — it has infinitely many solutions (an identity). But 4x + 1 = 4x + 9 simplifies to 1 = 9, always false — no solution. Recognizing these cases is as important as computing: after simplifying, if the variable vanishes, the truth of the remaining statement tells you the solution set.

Inequalities use <, >, ≤, ≥ instead of =. They solve almost exactly like equations — with one critical twist: multiplying or dividing by a negative number flips the inequality sign. Solving −2x < 10 gives x > −5. Why? On the number line, multiplying by −1 reflects everything across 0, reversing order. Solutions are intervals, graphed with open circles (for <, >) or closed circles (for ≤, ≥) and shading in the solution direction.

Try it yourself: Solve: (a) 3(2x − 1) = 5x + 8, (b) 7 − 2x = 3(x + 4), (c) solve and graph −3x + 4 ≥ 19.

Lesson 2: Absolute Value Equations and Inequalities

The absolute value |x| is a number's distance from zero on the number line, so it is never negative: |5| = 5 and |−5| = 5. An equation like |x| = 4 therefore has two solutions, x = 4 or x = −4, because both points sit 4 units from zero. For |x − 3| = 7, the expression inside is 7 units from zero: x − 3 = 7 or x − 3 = −7, giving x = 10 or x = −4. The pattern: isolate the absolute value first, then split into two equations.

Absolute value inequalities split by direction. |x| < 4 means x is within 4 of zero: −4 < x < 4, a single interval (AND). But |x| > 4 means x is far from zero in either direction: x < −4 or x > 4, two rays (OR). A memory hook: “less than” gives one interval (“less thAND”), “greater than” gives two rays (“greatOR”). The same rules apply after isolating: |2x + 1| ≤ 9 splits into −9 ≤ 2x + 1 ≤ 9, which solves to −5 ≤ x ≤ 4.

Absolute value models real tolerances. A machine that cuts rods to 50 cm with a tolerance of 0.2 cm accepts lengths satisfying |L − 50| ≤ 0.2, i.e. 49.8 ≤ L ≤ 50.2. A quality-control spec written as |x − target| ≤ tolerance is the standard way engineers express “close enough.”

Try it yourself: Solve: (a) |x + 5| = 3, (b) |x − 2| < 6, (c) |2x| ≥ 8. Answers: (a) x = −2 or x = −8; (b) −4 < x < 8; (c) x ≤ −4 or x ≥ 4.

Lesson 3: Word Problems — Turning Stories into Equations

Word problems are translation problems. The reliable routine is: (1) name the unknown with a variable, (2) translate each sentence into algebra, (3) solve, (4) check the answer against the story. Key translation phrases: “5 more than a number” is x + 5; “twice a number” is 2x; “5 less than a number” is x − 5 (order matters!); “the product of 3 and a number” is 3x; “is” almost always means equals.

Consecutive integer problems are a classic type: if the first integer is n, the next two are n + 1 and n + 2. “Three consecutive integers sum to 72” becomes n + (n + 1) + (n + 2) = 72, so 3n + 3 = 72, n = 23, and the integers are 23, 24, 25. Age problems use the same idea with a time shift: if Ana is now x and her brother is 3 years older, in 5 years their ages are x + 5 and x + 8. Money problems use value × quantity: 5 dimes and q quarters total 5(0.10) + 0.25q dollars.

Perimeter and geometry problems translate through formulas. A rectangle whose length is 4 more than its width, with perimeter 44, becomes: width w, length w + 4, so 2w + 2(w + 4) = 44, giving 4w + 8 = 44, w = 9, length 13. The final step is always a reality check: lengths positive, counts whole, answers matching what was asked (the width, not just w).

Try it yourself: (a) Two consecutive even integers sum to 58. Find them. (b) A rectangle's length is twice its width and its perimeter is 36. Find the width. Answers: (a) n + (n + 2) = 58, n = 28, so 28 and 30. (b) 2w + 2(2w) = 36, 6w = 36, w = 6.

Key vocabulary: Translate — converting words to algebraic expressions; consecutive integers — integers that follow one another, like 7, 8, 9.

Lesson 4: Functions and Graphing Lines

A function is a rule that assigns each input exactly one output. The notation f(x) = 2x + 1 names the function f and says: double the input, then add one. The domain is the set of allowed inputs; the range is the set of possible outputs. Not every equation is a function — the vertical line test decides: if any vertical line crosses a graph more than once, one input has multiple outputs, so it fails. x = y² fails; y = x² passes.

A linear function has the form f(x) = mx + b, and its graph is a straight line. The slope m measures steepness: rise over run, or (change in y)/(change in x). A positive slope climbs left to right; a negative slope falls; zero slope is horizontal; an undefined slope is vertical. The y-intercept b is where the line crosses the y-axis (x = 0). With m and b, graphing is instant: plot (0, b), then step by slope. The line y = −(3/2)x + 4 falls 3 units for every 2 it runs right.

Slope also appears between any two points: m = (y₂ −; y₁)/(x₂ −; x₁). Parallel lines share the same slope; perpendicular lines have slopes that are negative reciprocals (m₁ · m₂ = −1). Point-slope form, y − y₁ = m(x − x₁), builds a line's equation directly from a point and a slope — often faster than slope-intercept when the intercept isn't given.

Try it yourself: Find the equation (slope-intercept form) of the line through (−1, 4) with slope 3, and of the line through (0, −2) and (4, 6). Is y = 2x − 1 perpendicular to y = −x/2 + 7?

Lesson 5: Linear Models and Lines of Best Fit

Real data rarely falls exactly on a line, but a line often captures the trend. A scatter plot shows paired data as points; a line of best fit (trend line) is the straight line that passes as close as possible to all of them. Its slope is the average rate of change: if a line of best fit for study hours vs. test scores has slope 6, each extra hour of study is associated with about 6 extra points. Its y-intercept is the predicted value when the input is zero.

To build a rough line of best fit by hand, plot the points, draw a line that balances the points above and below it, then pick two points on your line (not necessarily data points) to compute its equation. For prediction, substitute into the equation: this is interpolation if the input lies inside the data's range, and extrapolation if it lies outside. Extrapolation is riskier — trends can bend. Predicting a 10-year-old's height from baby growth data would be nonsense.

A key caution: a strong linear pattern shows correlation, not causation. Ice cream sales and drownings rise together, but ice cream does not cause drowning — summer heat drives both. Always ask whether a third variable could explain the pattern before claiming one thing causes another.

Try it yourself: A fit line for height (in) vs. arm span (in) is y = 0.98x + 1.2. Predict the arm span of someone 65 in tall. Is this interpolation if the data spans 55–72 in? Answers: y = 0.98(65) + 1.2 = 64.9 in; yes, 65 is inside the range, so interpolation.

Key vocabulary: Scatter plot — data shown as points; line of best fit — the line closest to all data points; interpolation/extrapolation — predicting inside/outside the data range; correlation — a pattern of co-variation, not proof of cause.

Lesson 6: Graphing Linear Inequalities in Two Variables

A linear inequality in two variables, like y < 2x + 1, has a whole half-plane of solutions. The procedure: graph the boundary line y = 2x + 1 first. Use a dashed line for < or > (boundary not included) and a solid line for ≤ or ≥ (boundary included). Then shade one side. To decide which side, test a point not on the line — (0, 0) is the easiest when it is not on the boundary: 0 < 1 is true, so shade the side containing (0, 0).

Vertical and horizontal boundaries follow the same rules: x ≥ 2 is a solid vertical line with shading to the right; y < −1 is a dashed horizontal line with shading below. For inequalities in standard form, like 3x + 2y ≤ 12, testing (0, 0) gives 0 ≤ 12 (true), so shade toward the origin. A system of inequalities shades the overlap of the regions — the points satisfying every inequality at once. That overlap is the feasible region in optimization problems.

These graphs model constraints: a factory producing tables (t) and chairs (c) with wood limit 4t + 2c ≤ 100 and labor limit t + 3c ≤ 60. The feasible region's corner points are where the optimum (maximum profit) will be found — the seed of linear programming, used in logistics, manufacturing, and scheduling.

Try it yourself: Graph: (a) y < x − 2, (b) 2x + y ≥ 4. For (b), is (0, 0) shaded? Answers: (a) Dashed line through (0, −2) and (2, 0), shade below/right (test (0,0): 0 < −2 false, so shade away from origin). (b) Solid line, (0, 0) gives 0 ≥ 4 false, so not shaded.

Key vocabulary: Boundary line — the line dividing the plane into two half-planes; half-plane — all points on one side of a line; feasible region — the overlap of all constraints in a system.

Lesson 7: Systems of Linear Equations

A system of equations asks for the point(s) satisfying every equation at once. Graphically, each linear equation is a line, and the solution is their intersection. Two lines can intersect once (one solution, independent system), coincide (infinitely many solutions, dependent), or run parallel (no solution, inconsistent). The graph tells the story, but algebra gives exact answers.

The substitution method solves one equation for a variable and plugs that expression into the other — ideal when a variable already stands alone, as in y = 3x − 1. The elimination method adds or subtracts multiples of the equations to cancel a variable — ideal when coefficients line up, like 2x + 3y = 12 and 2x − y = 4 (subtract to eliminate x immediately). Choose the method the system invites; forcing the wrong one wastes time.

Systems model real trade-offs: two phone plans, two job offers, break-even analysis. If Plan A costs 30 + 0.10m dollars for m minutes and Plan B costs 0.25m, setting them equal (30 + 0.10m = 0.25m) gives m = 200 — the break-even point. Below 200 minutes Plan B wins; above, Plan A. The intersection point of the two cost lines is literally the moment the better deal switches.

Try it yourself: Solve by your choice of method: (a) y = 2x + 1 and y = −x + 7, (b) 4x + y = 10 and 2x + 3y = 12. Then decide: does the system x + y = 5, 2x + 2y = 10 have 0, 1, or infinitely many solutions?

Lesson 8: Systems in Context — Break-Even and Mixtures

Systems shine on problems where two quantities interact. A break-even problem compares cost and revenue: a business with fixed costs $1,200 and $15 per unit, selling at $35 per unit, has cost C = 1200 + 15n and revenue R = 35n. Setting them equal, 1200 + 15n = 35n, gives 20n = 1200, n = 60 units. Below 60 units the business loses money; above it, profit begins. The break-even point is the intersection of the two lines.

Mixture problems pair a total-amount equation with a total-value equation. How many liters of 20% acid solution and 50% acid solution make 10 L of 32% solution? Let x and y be the liters: x + y = 10 (amount) and 0.20x + 0.50y = 0.32(10) = 3.2 (acid). Substituting y = 10 − x gives 0.20x + 0.50(10 − x) = 3.2, so 0.20x + 5 − 0.50x = 3.2, −0.30x = −1.8, x = 6 L of 20% and y = 4 L of 50%. The same two-equation template solves ticket, coin, and investment problems.

Distance-rate-time problems with two travelers also become systems: d = rt for each traveler, with a shared distance or time linking them. A boat going 12 mph downstream and 4 mph upstream over the same 24-mile course takes 2 hours down and 6 hours back — the current's effect appears as +c and −c in the two speeds.

Try it yourself: A shop sells pens at $2 and notebooks at $5. 40 items sold for $125. How many notebooks? Answers: p + n = 40, 2p + 5n = 125 → 2(40 − n) + 5n = 125 → 3n = 45, n = 15 notebooks.

Key vocabulary: Break-even point — where cost equals revenue; mixture problem — combining two concentrations/values to a target; system in context — equations modeling a real situation.

Lesson 9: Polynomials and Factoring

A polynomial is a sum of terms, each a constant times a variable to a whole-number power: 4x³ − 2x + 7. The degree is the highest exponent (here, 3). Polynomials are classified by term count: monomial (one term), binomial (two), trinomial (three). Adding and subtracting is just combining like terms; multiplying uses the distributive property repeatedly — FOIL (First, Outer, Inner, Last) organizes the product of two binomials: (x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15.

Factoring reverses multiplication: writing a polynomial as a product of simpler ones. Always start with the greatest common factor (GCF): 6x³ + 9x² = 3x²(2x + 3). For trinomials x² + bx + c, find two numbers with product c and sum b: x² + 7x + 12 = (x + 3)(x + 4) because 3 · 4 = 12 and 3 + 4 = 7. For ax² + bx + c with a ≠ 1, try factor pairs of a and c systematically (or the ac-method).

Memorize the special patterns: difference of squares, a² − b² = (a − b)(a + b); perfect square trinomials, a² ± 2ab + b² = (a ± b)². So x² − 49 = (x − 7)(x + 7), and 9x² + 12x + 4 = (3x + 2)². Factoring is the key that unlocks quadratic equations in the next lesson: if a product equals zero, one of the factors must be zero.

Try it yourself: Factor: (a) x² − 11x + 24, (b) 4x² − 25, (c) 5x³ − 20x, (d) x² + 6x + 9.

Lesson 10: Introduction to Quadratics

A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0, and its graph is a parabola — the U-shape of a thrown ball's path or a satellite dish. The parabola's lowest (or highest) point is the vertex; the vertical line through it is the axis of symmetry. If a > 0 the parabola opens upward (vertex is a minimum); if a < 0 it opens downward (vertex is a maximum).

The simplest solution method is factoring + the Zero Product Property: if AB = 0, then A = 0 or B = 0. To solve x² + 5x + 6 = 0, factor to (x + 2)(x + 3) = 0, then x = −2 or x = −3. Each factor gives one solution — quadratics can have up to two. When factoring fails (or is ugly), the quadratic formula, x = (−b ± √(b² − 4ac)) / 2a, always works: it was derived by completing the square on the general quadratic.

The discriminant, b² − 4ac (the part under the square root), predicts the solution count before you compute: positive → two real solutions; zero → one repeated solution; negative → no real solutions (the parabola never touches the x-axis). For x² − 6x + 9 = 0, the discriminant is 36 − 36 = 0, so x = 3 is a double root — the vertex sits exactly on the x-axis.

Try it yourself: Solve by factoring: (a) x² − 9x + 20 = 0, (b) 3x² + 10x + 8 = 0. Use the quadratic formula on x² + 4x + 2 = 0 and state how many real solutions x² + x + 5 = 0 has.

Lesson 11: Graphing Quadratics — Vertex Form

Every parabola can be written in vertex form, y = a(x − h)² + k, where (h, k) is the vertex. The number a controls the shape: |a| > 1 makes the parabola narrower, 0 < |a| < 1 makes it wider, and a < 0 flips it upside down. The graph of y = 2(x − 3)² + 1 has vertex (3, 1), opens upward, and is narrower than y = x². Vertex form is the fastest way to sketch a parabola: plot the vertex, then use a to step away.

To convert standard form y = ax² + bx + c into vertex form, use completing the square. For y = x² + 6x + 5: take half of 6 (which is 3), square it (9), and add/subtract it inside: y = (x² + 6x + 9) + 5 − 9 = (x + 3)² − 4. Vertex (−3, −4). The axis of symmetry is the vertical line x = h, here x = −3. The vertex is also the maximum or minimum value of the function — the launch height of a ball, the maximum profit, the minimum cost.

The x-intercepts (roots) are found by setting y = 0. For y = (x + 3)² − 4, solving (x + 3)² = 4 gives x + 3 = ±2, so x = −1 or x = −5. The y-intercept is found by setting x = 0: y = 5. With vertex, intercepts, and the direction of opening, a parabola sketch is complete and accurate.

Try it yourself: Find the vertex and roots of y = (x + 1)² − 9. Is the vertex a max or min? Answers: Vertex (−1, −9); roots: (x+1)² = 9 → x = 2 or x = −4; a = 1 > 0, so a minimum.

Key vocabulary: Vertex form — y = a(x − h)² + k; completing the square — rewriting a quadratic as a perfect square plus a constant; axis of symmetry — the vertical line through the vertex.

Lesson 12: Exponential Functions and Growth/Decay

An exponential function has the form y = a · b^x, where a is the starting value and b is the growth factor. Unlike a linear function, which adds the same amount each step, an exponential function multiplies by the same factor each step. If b > 1, the function grows; if 0 < b < 1, it decays. The graph curves upward for growth, flattens toward a horizontal asymptote (usually y = 0) for decay, and always passes through (0, a).

Percent change converts directly to b: a 5% increase per year means b = 1.05; a 20% decrease means b = 0.80. Compound interest, population growth, and radioactive decay are all exponential. The doubling time is how long it takes a growing quantity to double; the half-life is how long it takes a decaying quantity to halve. Both come from solving a · b^t = 2a or a/2.

Real-world application: Banks pay compound interest exponentially; a $1,000 deposit at 4% annual interest becomes 1000(1.04)^t. Viruses, rumors, and investments spread exponentially too, which is why early growth looks slow and then explodes.

Try it yourself: (a) A bacteria culture starts with 500 cells and doubles every hour. How many after 6 hours? (b) A $2,000 laptop depreciates 20% per year. What is it worth after 3 years? Answers: (a) 500 · 2⁶ = 32,000; (b) 2000(0.8)³ = $1,024.

Key vocabulary: Exponential function — y = a · b^x, multiplying by b each step; growth/decay factor — b, greater than 1 for growth, between 0 and 1 for decay; asymptote — a line a graph approaches but never touches.

Lesson 13: Radicals and Square Roots

A radical is a root symbol: √x is the square root, ∛x the cube root. The number inside is the radicand. To simplify a square root, factor the radicand and pull out pairs: √50 = √(25 · 2) = 5√2. A radical is in simplest form when the radicand has no perfect-square factors (other than 1) and no fractions. The product rule √a · √b = √ab and quotient rule √(a/b) = √a/√b justify every simplification.

Radicals combine like like terms: 3√5 + 2√5 = 5√5, but √2 + √3 cannot be combined. Multiplying uses the distributive property: √2(3 + √5) = 3√2 + √10. Rationalizing the denominator removes a radical from a denominator by multiplying by a form of 1: 1/√3 becomes √3/3. For a binomial denominator like 1/(2 + √3), multiply by its conjugate, 2 − √3.

Real-world application: Carpenters use square roots when a diagonal is involved (√2 appears in every square's diagonal). Engineers rationalize denominators so measurements divide cleanly, and the Pythagorean Theorem constantly produces radicals that must be simplified.

Try it yourself: Simplify: (a) √98, (b) 4√3 − √3, (c) rationalize 2/√5. Answers: (a) 7√2; (b) 3√3; (c) 2√5/5.

Key vocabulary: Radical — a root symbol; radicand — the number under the radical; simplest form — no perfect-square factors in the radicand, no radical in a denominator; conjugate — a + b√c paired with a − b√c.

Lesson 14: Rational Expressions

A rational expression is a fraction whose numerator and denominator are polynomials, like (x² − 9)/(x + 3). Simplifying works exactly like simplifying a numerical fraction, except you factor first and cancel common factors — never terms. (x² − 9)/(x + 3) = (x − 3)(x + 3)/(x + 3) = x − 3. The values that make the original denominator zero are excluded values: here x ≠ −3, and that restriction survives simplification, because the original expression was undefined there.

Multiplying rational expressions: factor everything, then cancel. Dividing: flip the second expression and multiply. Adding and subtracting: find the least common denominator (LCD), rewrite each expression with it, then combine numerators. The LCD is the product of each distinct factor at its highest power. A common error is canceling terms instead of factors: (x + 2)/x does NOT simplify to 2.

Real-world application: Rational expressions model rates and work: if one hose fills a pool in x hours and another in y hours, their combined rate is 1/x + 1/y. Average speed over a round trip, lens formulas, and electrical resistance all produce rational expressions.

Try it yourself: Simplify: (a) (2x + 6)/(x² − 9), (b) (x/(x+2)) · ((x² − 4)/x). State excluded values. Answers: (a) 2/(x − 3), x ≠ ±3; (b) x − 2, x ≠ 0, −2, 2.

Key vocabulary: Rational expression — a ratio of polynomials; excluded value — a value making a denominator zero; least common denominator — the smallest expression divisible by all denominators.

Lesson 15: Solving Rational and Radical Equations

A rational equation contains a variable in a denominator. The standard method: multiply every term by the LCD to clear the fractions, then solve the resulting polynomial equation. A radical equation contains a variable under a radical. The method: isolate the radical, raise both sides to the matching power, solve, and — critically — check every answer in the original equation.

Checking is not optional, because both methods can create extraneous solutions: values that satisfy the cleared equation but not the original. Squaring both sides can turn −3 into +3 (both square to 9), and multiplying by the LCD can introduce a value that makes a denominator zero. An answer that makes any denominator zero, or that fails the original equation, must be rejected.

Real-world application: Work-rate problems (two people painting a house together), distance problems with fractional speeds, and pendulum-period formulas all lead to rational or radical equations. Engineers solving for a cable length from a sag formula must reject the extraneous root that gives a negative length.

Try it yourself: Solve and check: (a) 2/x + 1 = 5/x, (b) √(x + 5) = x − 1. Answers: (a) Multiply by x: 2 + x = 5, x = 3 (valid). (b) Square: x + 5 = x² − 2x + 1, x² − 3x − 4 = 0, x = 4 or x = −1; check: x = 4 works, x = −1 fails (3 ≠ −2). Only x = 4.

Key vocabulary: Extraneous solution — a value produced by solving that does not satisfy the original equation; cross-multiplication — solving a/b = c/d as ad = bc.

Lesson 16: Statistics: Data Displays, Two-Way Tables, and Correlation

Statistics turns raw data into decisions. A box plot summarizes a data set with five numbers: minimum, first quartile (Q1), median, third quartile (Q3), and maximum. The box spans Q1 to Q3 (the middle 50% of the data), and the whiskers reach the extremes. The interquartile range, IQR = Q3 − Q1, measures spread and is resistant to outliers, unlike the range.

A two-way table organizes counts by two categorical variables, like grade level vs. whether a student plays a sport. Relative frequencies (each count divided by its row, column, or grand total) reveal patterns: if 70% of juniors but only 40% of freshmen play sports, grade and sports participation are associated. For numerical data, a scatter plot's correlation measures the strength and direction of a linear pattern — but correlation never proves causation.

Real-world application: Hospitals use box plots to compare recovery times; schools use two-way tables to study course enrollment patterns; businesses track whether ad spending correlates with sales before deciding whether the ads actually cause the sales.

Try it yourself: Data: 4, 8, 9, 10, 12, 15, 20. Find the median, Q1, Q3, and IQR. Answers: Median = 10; Q1 = 8; Q3 = 15; IQR = 7.

Key vocabulary: Box plot — a five-number-summary display; quartile — a quarter-point of ordered data; two-way table — counts by two categorical variables; relative frequency — a count as a fraction of a total.

Lesson 17: Systems of Inequalities and Linear Programming Preview

A system of inequalities has a solution set that is the overlap of all its half-planes. Graph each inequality (dashed for </>, solid for ≤/≥, shade the correct side), and the feasible region is where all the shadings overlap. Points in the feasible region satisfy every constraint at once; points outside fail at least one. The corners (vertices) of the feasible region are the most important points: in optimization, the best value always occurs at a corner.

Linear programming is the method of maximizing or minimizing a linear objective function (like profit P = 40x + 30y) subject to linear constraints. The procedure: graph the constraints, find the feasible region's vertices, and evaluate the objective function at each vertex. The largest (or smallest) value is the optimum. This is how factories, airlines, and farms make allocation decisions.

Real-world application: A bakery deciding how many cakes and cookies to bake, given oven time and ingredient limits, solves a linear program. A shipping company minimizing cost subject to delivery requirements does the same, with thousands of variables and computer solvers.

Try it yourself: Maximize P = 2x + 3y subject to x + y ≤ 5, x ≥ 0, y ≥ 0. Answer: Vertices (0,0), (5,0), (0,5); P values 0, 10, 15; maximum 15 at (0, 5).

Key vocabulary: Feasible region — points satisfying all constraints; constraints — the inequalities limiting a problem; objective function — the quantity to maximize or minimize.

Lesson 18: Quadratic Applications: Projectile Motion and Optimization

Quadratic functions model anything that rises then falls, or falls then rises. The height of a projectile is h(t) = −16t² + v₀t + h₀ (feet and seconds), where v₀ is the initial upward velocity and h₀ the launch height. The coefficient −16 comes from gravity. The vertex gives the maximum height and when it occurs: t = −b/(2a). The zeros give when the object hits the ground (h = 0); the positive zero is the flight time.

Quadratics also optimize area and profit. A farmer with 200 feet of fencing who wants the largest rectangular pen maximizes A = x(100 − x) = −x² + 100x, whose vertex is at x = 50 — a 50-by-50 square, area 2500 ft². A business whose profit is P(x) = −2x² + 120x − 500 maximizes profit at the vertex, x = 30. In every case, the vertex formula turns a story into one division.

Real-world application: Engineers design water fountains so the arc lands in the basin; farmers and builders maximize area with limited fencing; businesses price products by maximizing a quadratic profit model. Every parabola vertex is a best or worst case.

Try it yourself: h(t) = −16t² + 32t + 5. Find the maximum height and the time it occurs. Answers: t = 1 s; h(1) = −16 + 32 + 5 = 21 ft.

Key vocabulary: Projectile — an object thrown or launched under gravity; vertex — the maximum or minimum point; axis of symmetry — the vertical line x = −b/(2a).

Lesson 19: Sequences: Arithmetic and Geometric

A sequence is an ordered list of numbers. In an arithmetic sequence, each term adds a constant common difference d: 5, 9, 13, 17, ... has d = 4. The explicit formula is a_n = a₁ + (n − 1)d. In a geometric sequence, each term multiplies by a constant common ratio r: 3, 6, 12, 24, ... has r = 2. Its explicit formula is a_n = a₁ · rⁿ⁻¹. A recursive formula instead defines each term from the previous one: a_n = a_n−1 + d.

The sum of the first n terms has closed forms. For arithmetic: S_n = n/2 · (a₁ + a_n), or equivalently n/2 · (2a₁ + (n − 1)d). For geometric: S_n = a₁(1 − rⁿ)/(1 − r) for r ≠ 1. Gauss famously summed 1 + 2 + ... + 100 by pairing 1 + 100, 2 + 99, ..., fifty pairs of 101, giving 5050 — exactly the arithmetic-sum formula.

Real-world application: Savings with regular deposits form an arithmetic sequence; compound interest and viral spread form geometric sequences. Computer scientists analyze how sequences grow to predict how algorithms slow down on large inputs.

Try it yourself: (a) Find the 15th term of 100, 93, 86, ... (b) Find the 6th term of 5, 10, 20, ... Answers: (a) d = −7, a₁₅ = 100 + 14(−7) = 2; (b) r = 2, a₆ = 5 · 2⁵ = 160.

Key vocabulary: Sequence — an ordered list of numbers; common difference/ratio — the constant added / multiplied; explicit formula — a_n directly in terms of n; recursive — each term defined from the previous one.

Lesson 20: Algebra I Capstone: Modeling Real Problems

Real problems do not arrive labeled by chapter. Modeling means choosing the right kind of function for the situation: linear when a quantity changes by a constant amount per step, quadratic when it rises then falls or involves area, exponential when it multiplies by a constant factor per step. The modeling cycle is: understand the situation, choose variables and a model, solve, then validate — check the answer against the real world and refine the model if it fails.

Consider a food truck: fixed costs $300 per day, $4 per meal, selling at $11 per meal. Profit P(n) = 11n − (300 + 4n) = 7n − 300. Break-even: 7n = 300, n ≈ 43 meals. A second question: if the truck lowers the price to $9, demand rises to 80 meals; at $11, demand is 60. Assume demand is linear in price: slope = (60 − 80)/(11 − 9) = −10 meals per dollar. Which price earns more? At $11: revenue 660; at $9: revenue 720 — but profit also subtracts costs, so the full model decides. Capstone thinking means carrying the model all the way to the decision.

Real-world application: Choosing a phone plan, pricing a product, sizing a garden, and predicting a population are all capstone-style problems: no one tells you which chapter to use. The skill is recognizing the pattern — constant change, squared relationship, or constant multiplier — and validating the result.

Try it yourself: A taxi charges $3 pickup plus $2 per mile. A second charges $1 pickup plus $2.50 per mile. At how many miles do they cost the same? Which is cheaper for a 12-mile ride? Answers: 3 + 2m = 1 + 2.5m gives m = 4 miles; at 12 miles, first is $27, second is $31, so the first is cheaper.

Key vocabulary: Model — a mathematical description of a real situation; assumption — a simplification made to build a model; validation — checking a model's predictions against reality.