Unit 14 takes the function families you've already learned (linear, quadratic, exponential, logarithmic, rational, trig) and puts them to work on physics, business, money, and data problems. By the end, you should be able to turn a word problem into an equation, solve it, and explain what the answer means in the situation.
Read the overview first, then work through each topic note and try the check-yourself question before you look at the answer.
This is an applications unit, so there are very few brand-new formulas. The new skill is translation, turning sentences into equations and turning numerical answers back into sentences. Schools vary in how much physics or finance they include, but the modeling process is the same everywhere.
Honors Algebra II Unit 14 covers problem-solving with real-world applications. You model situations with functions and interpret their parameters (14.1). You use quadratics and other equations for motion, projectiles, and forces in physics and engineering (14.2). You find maximums and minimums using the vertex of a parabola and the corner points of a feasible region (14.3). You apply exponential and logarithmic functions to compound interest, loans, and annuities, and use regression to fit models to data (14.4).
For equally spaced inputs, a constant difference between outputs suggests a linear model, and a constant ratio suggests an exponential one. A rise-then-fall shape suggests a quadratic, and a repeating pattern suggests a trig function. Picking the family is the biggest decision in any model.
In , the 25 is a starting fee and the 0.50 is a cost per mile. In , the 64 is the launch height in feet. If you can't say what a parameter means, you don't fully understand the model yet.
Math often produces solutions that make no sense in context, like a negative time or 2.7 workers. You keep only the solutions that fit the domain of the situation, and you round in the direction the situation requires.
Algebra skills from earlier units handle the solving step. This unit adds the steps on either side, setting up the right equation from a situation and explaining the result in plain language. Those two steps are what AP Precalculus and calculus will expect you to handle on your own.
A mathematical model represents a real situation using mathematical relationships, such as an equation relating two quantities.
Physics formulas are just equations with real units, so the algebra you know applies directly.
Optimization means finding the input that makes some quantity as large or as small as possible.
Simple interest grows linearly, I = Prt, because you only earn on the original principal.
Open this guide for a closer review of the topic.
A mathematical model represents a real situation using mathematical relationships, such as an equation relating two quantities. First define your variables with units, like = hours and = population. Then look at how the data changes and pick a function family. Finally, use given values or regression to find the parameters.
A useful test uses equally spaced inputs. Constant differences suggest a linear model, a constant multiplication factor suggests an exponential model, and constant nonzero second differences suggest a quadratic model. A finite table alone does not prove which model describes the situation.
Some situations change rules partway through, like shipping that costs a flat 1.50 per extra pound. That calls for a piecewise function.
Worked example. A bacteria culture starts at 500 cells and doubles every 3 hours. The starting value is and the growth factor is 2 per 3 hours, so . After 12 hours, cells. To find when the count reaches 20,000, solve , so and hours.
A table shows outputs 3, 6, 12, 24 for inputs 0, 1, 2, 3. Which model fits, and what is its equation? The model is exponential because each output is multiplied by 2, so .
| Pattern in the data | Function family | Example situation |
|---|---|---|
| Constant difference | Linear, $y = mx + b$ | Rental with a base fee plus a per-mile charge |
| Constant ratio for equally spaced inputs | Exponential, $y = ab^x$ | Population doubling every few hours |
| Rises then falls, with constant second differences for equally spaced inputs | Quadratic, $y = ax^2 + bx + c$ | Height of a thrown ball |
| An increase that becomes slower as the input grows | Logarithmic, $y = a\ln x + b$ | Earthquake magnitude vs. amplitude |
| Repeats on a cycle | Trigonometric, $y = a\sin(bx) + d$ | Ferris wheel rider height over time |
Physics formulas are just equations with real units, so the algebra you know applies directly. One common formula is the projectile height model . It assumes constant gravity near Earth's surface and ignores air resistance. With in seconds, use ft/s² for height in feet or m/s² for height in meters. Here is the starting upward velocity and is the starting height.
Other formulas show up as literal equations you rearrange. Newton's second law gives , so a 36 N net force on a 12 kg box gives m/s². Kinetic energy means doubling speed multiplies energy by 4, because of the square.
Worked example. A ball is thrown up from a 64-foot roof at 48 ft/s, so . To find when it lands, set and factor out to get . Then , so or . Time can't be negative, so the ball lands after 4 seconds.
A rock is dropped from 44.1 m with no initial velocity. When does it hit the ground? Set , so and seconds.
Optimization means finding the input that makes some quantity as large or as small as possible. The quantity you care about is the objective function, and the limits on your choices are the constraints. In Algebra II, you handle these without calculus in two main ways.
For a quadratic objective, the vertex is at . On the full real line, a negative gives a maximum there and a positive gives a minimum. With constraints, check whether the vertex is allowed and compare it with any applicable boundary values. For the ball in 14.2, the peak is at s, at a height of 100 ft.
When the objective and constraints are all linear, you're doing linear programming. For a bounded feasible region with corners, graph the constraints and evaluate the objective at each corner. If a best value exists, at least one corner gives it, though points along an edge may also give the same value. Other linear programs can have no feasible solution or no finite optimum.
Worked example. You have 240 ft of fence for three sides of a rectangle along a river. Let be each side perpendicular to the river, so the parallel side is . Area is . The vertex is at , giving sides of 60 ft and 120 ft and a maximum area of 7,200 ft².
Profit is and the feasible region has corners (0, 0), (0, 10), (6, 6), and (8, 0). What is the maximum profit? The corners give 0, 40, 54, and 40, so the maximum is 54 at (6, 6).
Simple interest grows linearly, , because you only earn on the original principal. Compound interest grows exponentially, , because you earn interest on past interest. Continuous compounding uses . For 2,240, and monthly compounding gives about $2,254.54.
To solve for time, you need logarithms. The Rule of 72 gives a fast estimate of doubling time, which is 72 divided by the percent rate.
For equal payments made at the end of each period, the future value of an annuity is , where is the nonzero rate per period and is the number of payments. If , the future value is . For a typical fixed-rate loan paid off with equal payments, the interest portion decreases and the principal portion increases over time.
On the data side, regression fits a model to points. For a linear relationship, measures the strength and direction of the association, and describes the proportion of variation in the output explained by the fitted line. A strong fit still doesn't prove that one variable causes the other.
Worked example. How long does 2 = (1.015)^{4t}4t = \frac{\ln 2}{\ln 1.015} \approx 46.672 \div 6 = 12$ years, which is close.
Which earns more in 10 years on 1000(1.05)^{10} \approx $1{,}628.891,500 for simple interest.
| Type | Formula | Growth pattern |
|---|---|---|
| Simple interest | $A = P(1 + rt)$ | Linear |
| Compound interest | $A = P\left(1 + \frac{r}{n}\right)^{nt}$ | Exponential, in steps |
| Continuous compounding | $A = Pe^{rt}$ | Exponential and continuous in time |
| Term | Definition |
|---|---|
| Mathematical model | An equation or function that represents how quantities in a real situation relate to each other. |
| Parameter | A fixed constant in a model, like a starting value or growth rate, that has a real-world meaning. |
| Growth factor | The base $b$ in $y = ab^x$, which tells you what the output is multiplied by for each one-unit increase in input. |
| Vertex | The turning point of a parabola, located at $x = -\frac{b}{2a}$, where a quadratic reaches its maximum or minimum. |
| Constraint | A limit on the variables in a problem, usually written as an equation or inequality. |
| Linear programming | A method for optimizing a linear objective subject to linear constraints. For a bounded feasible region with corners, checking those corners finds an optimum. |
| Principal | The original amount of money invested or borrowed. |
| Amortization | Paying down a loan over time through payments that reduce the principal and cover any interest due. |
| Residual | The difference between an actual data value and the value a regression model predicts. |
Annuity and loan formulas use the rate per period and the number of periods. For a 6% loan paid monthly over 5 years, use and , not 0.06 and 5.
A quadratic can have zero, one, or two real solutions, and a negative time or length may not fit the situation. Always check each solution against the real-world domain before you report it.
The vertex formula gives the input where the max or min happens. If the question asks for the greatest area or height, plug that input back into the function to get the actual value.
An close to 1 or -1 means the points follow a linear pattern closely. Two quantities can rise together because a third factor drives both, so describe the relationship as an association unless the situation gives a reason for cause.