Non Linear Functions Pattern - Create Equations

Creating Nonlinear Equations from Word Problems

This pattern gives you a real-world scenario and asks you to build the equation that models it. The result is usually a quadratic — often from multiplying two related quantities (price $\times$ quantity, length $\times$ width) or from using vertex form to model a maximum or minimum.

The Core Setups

Product of two related quantities: Revenue $=$ price $\times$ quantity sold. If quantity decreases as price increases ($Q = 300 - p$), then $R(p) = p(300 - p) = 300p - p^2$.

Area from related dimensions: If length $= w - 50$ and width $= w$, then $A = w(w - 50) = w^2 - 50w$.

Vertex form: If you know the maximum/minimum value and where it occurs: $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex.

Worked Examples

Example 1. The function $h$ gives the product of a number $y$ and a value that is $15$ less than $y$. Which equation defines $h$?

A) $h(y) = y^2 - 15y$
B) $h(y) = y^2 - 15$
C) $h(y) = 15y^2$
D) $h(y) = y^2 + 15y$

"A number $y$" → first factor is $y$.
"$15$ less than $y$" → second factor is $y - 15$.
Product: $y(y - 15) = y^2 - 15y$.
Gotcha: Option B subtracts $15$ as a constant instead of multiplying. "Product" means multiplication: $y \times (y - 15)$.
The answer is A.

Example 2. A rectangular garden has width $w$ meters and length that is $50$ meters less than the width. Which equation gives the area?

A) $A(w) = w^2 - 50$
B) $A(w) = w^2 - 50w$
C) $A(w) = w^2 + 50w$
D) $A(w) = 50w - w^2$

Width $= w$. Length $= w - 50$.
Area $= w \times (w - 50) = w^2 - 50w$.
Gotcha: Option A subtracts $50$ instead of $50w$. When you distribute $w$ across $(w - 50)$, you get $w^2 - 50w$, not $w^2 - 50$.
The answer is B.

Example 3. Revenue is the product of price $p$ and quantity $(300 - p)$. Which equation gives revenue?

$R(p) = p(300 - p) = 300p - p^2$
The answer choices may rearrange this as $R(p) = -p^2 + 300p$ or present it in different forms. Look for the version that matches $p \times (300 - p)$.

Example 4. A company's profit is maximized at $80,000 when the price is $25. At a price of $5, the profit is $0. Which equation models this?

Use vertex form: $P(x) = a(x - 25)^2 + 80$, where $P$ is in thousands.
Plug in the zero-profit point: $0 = a(5 - 25)^2 + 80 = a(400) + 80$
$a = -\dfrac{80}{400} = -0.2$
$P(x) = -0.2(x - 25)^2 + 80$
Gotcha: The vertex $(25, 80)$ gives you $h$ and $k$. The second point gives you $a$. Don't confuse the vertex with the zero or the intercept.

Example 5. A ball is thrown from the top of a building. Its height $h(t) = -16t^2 + v_0 t + h_0$, where $v_0$ is initial velocity and $h_0$ is initial height. If the building is $48$ feet tall and the ball is thrown upward at $32$ ft/s, write the equation.

$h(t) = -16t^2 + 32t + 48$
Key: $-16$ is the gravity constant (always the same for feet/seconds), $v_0 = 32$ is the launch speed, and $h_0 = 48$ is the starting height.

What to Do on Test Day

• "Product of" = multiply. If the problem says "the product of $x$ and something," write $x \times (\text{something})$ and distribute.
• "$k$ less than $x$" = $x - k$, not $k - x$. Read the order carefully.
• Vertex form: $f(x) = a(x-h)^2 + k$. Use the vertex to fill in $h$ and $k$, then use a second point to find $a$.
• Distribute carefully. $w(w - 50) = w^2 - 50w$, not $w^2 - 50$.
• Look for hidden quadratics. Any time two quantities that depend on the same variable are multiplied, you get a quadratic: $p \times (300 - p) = 300p - p^2$.