Turn Word Problems & Graphs into Clean Quadratic Models

A lot of SAT Math “hard” questions are really just stories or graphs hiding a quadratic.
If you can turn those into a clean equation like y = ax² + bx + c or
y = a(x – h)² + k quickly, they stop being scary and start feeling routine.

1. Big Idea: Stories and Graphs Are Just Quadratics in Disguise

On the SAT, quadratics don’t only appear as “Solve 2x² – 3x – 5 = 0.” They often show up as:

Height of a ball or rocket over time.
Profit, revenue, or area depending on some decision (price, length, etc.).
Graphs of parabolas where you’re asked about roots, vertex, or values.
Tables of values that secretly follow a quadratic pattern.

Goal of this page: learn a repeatable process to go from
“wordy story or graph” → “clean quadratic model” you can actually use.

2. Word Problems: Step-by-Step Model Setup

Almost every quadratic word problem can be attacked with the same four steps:

Choose your variable clearly.
Example: “Let x be the number of tickets,” or “Let t be the time in seconds.”
Write expressions for pieces that change.
Revenue = (price per item) × (number of items), Area = (length) × (width), Height = initial + change.
Combine into one quadratic expression.
Expand or simplify until you have something like ax² + bx + c.
Decide what the question wants.
Roots? Maximum? Value at a certain x? That tells you which quadratic method to use.

Mini Example: Ticket Revenue
A concert hall has 500 seats. Tickets are sold for $20. The manager finds that for every $1 increase
in price, 10 fewer tickets are sold. Write a quadratic model for revenue R(x) in terms of x, the number
of $1 increases.

Variable: Let x = number of $1 increases.
New price: 20 + x.
New number of tickets: 500 – 10x.
Revenue: R(x) = (price)(tickets) = (20 + x)(500 – 10x).
Expand:

R(x) = 20(500 – 10x) + x(500 – 10x)

= 10,000 – 200x + 500x – 10x²

= -10x² + 300x + 10,000.

Now you have a clean quadratic model: R(x) = -10x² + 300x + 10,000.

From here, you can find the price that gives maximum revenue (vertex), or check specific values (plug in x).

3. Vertex-Based Models (Max/Min Stories)

Many SAT word problems are clearly about a maximum or minimum:
maximum height, maximum profit, minimum cost, etc. Those are perfect for vertex form
or for using the vertex formula on standard form.

Vertex form:
y = a(x – h)² + k
Vertex: (h, k).
If a > 0 → minimum at y = k; if a < 0 → maximum at y = k.

Example 1: Projectile / Height Model

A ball is thrown upward from a platform. Its height, in meters, t seconds after it is thrown is modeled by:

h(t) = -5t² + 20t + 3.

What is the maximum height of the ball?
After how many seconds does it reach that height?

Show explanation

We already have a standard-form quadratic. Use the vertex:

Step 1: Find time at vertex.
For y = at² + bt + c, vertex t = -b / (2a).

Here, a = -5, b = 20:

t = -20 / (2 · -5) = -20 / -10 = 2.

Step 2: Find maximum height.
Plug t = 2 into h(t):

h(2) = -5(4) + 20(2) + 3 = -20 + 40 + 3 = 23.

So the ball reaches a maximum height of 23 meters at t = 2 seconds.

If you wanted vertex form, it would be h(t) = -5(t – 2)² + 23, but you don’t
actually need to rewrite it to answer the questions.

Example 2: Revenue / Profit Model

Using the ticket model from earlier:

R(x) = -10x² + 300x + 10,000

where x is the number of $1 increases. For what value of x is the revenue maximized, and what is that maximum revenue?

Show explanation

This is a “maximum revenue” question → use the vertex.

Step 1: Find x at vertex.
a = -10, b = 300.

x = -b / (2a) = -300 / (2 · -10) = -300 / -20 = 15.

So revenue is maximized when x = 15 (price is increased $15).

Step 2: Find maximum revenue.
R(15) = -10(15²) + 300(15) + 10,000

= -10(225) + 4500 + 10,000

= -2250 + 14,500

= 12,250.

Maximum revenue is $12,250 at 15 increases.

4. Root-Based & Factored Models

Sometimes the story tells you directly where the value is zero (starts at ground level, returns to ground,
no profit at certain prices, etc.). That’s perfect for building a model using the roots first
and writing the function in factored form.

Example 3: Height = 0 at Two Times

A ball is thrown upward from ground level. It leaves the ground at time t = 0, reaches a maximum height,
and then returns to the ground after 6 seconds. The height h(t) in meters is modeled by a quadratic function.
Assuming the graph is symmetric, write a possible function for h(t).

Show explanation

We know when the height is zero:

At t = 0 (start at ground).
At t = 6 (returns to ground).

These are roots of the height function. So factored form looks like:

h(t) = a·t(t – 6).

We don’t know the exact maximum height, so we can pick a negative a (because it opens down). For example:

h(t) = -2t(t – 6)

is a valid model that has roots at t = 0 and t = 6 and opens downward. If the problem gave another point
(like the maximum height), we’d plug it in to solve for a exactly.

Example 4: Profit is Zero at Two Prices

A company sells a product at price p dollars. Its profit P(p), in dollars, is zero when p = 5 and p = 25.
Profit is positive between these prices and negative outside them. Write a quadratic model for P(p).

Show explanation

We know the prices where profit is zero: p = 5 and p = 25. That means:

P(p) = a(p – 5)(p – 25)

Profit is positive between 5 and 25 and negative outside. That means the parabola opens downward
(a < 0). A simple choice is:

P(p) = -1 · (p – 5)(p – 25), or just
P(p) = -(p – 5)(p – 25).

If needed, you can expand:

(p – 5)(p – 25) = p² – 25p – 5p + 125 = p² – 30p + 125

So P(p) = -p² + 30p – 125.

5. From Graphs & Tables to Equations

The SAT also gives you graphs or tables of a parabola and asks you to identify or interpret the equation.
This is where knowing what each form shows you becomes powerful.

A. Building from a Graph

From a graph of a parabola, you should look for:

The x-intercepts (for factored form).
The vertex (for vertex form).
The y-intercept (for standard form’s c).
Whether the graph opens up or down, and whether it’s “wide” or “narrow” (value/sign of a).

Example 5: Graph with Visible Vertex and Root

Suppose a graph shows a parabola that:

Has a vertex at (2, 3).
Passes through the point (0, -1).
Opens upward.

Find an equation in standard form.

Show explanation

Start with vertex form because we know the vertex:

y = a(x – 2)² + 3.

Use the point (0, -1) to solve for a:

-1 = a(0 – 2)² + 3

-1 = a(4) + 3

-4 = 4a

a = -1.

But the graph was said to open upward — that means our data (or assumption) must have had
a mistake, or the SAT would give consistent info. Let’s fix the “opens upward” detail:

If the graph actually opens downward, then a = -1 is correct, and the equation is:

y = -(x – 2)² + 3.

Expand to standard form:

y = -(x² – 4x + 4) + 3

y = -x² + 4x – 4 + 3

y = -x² + 4x – 1.

B. Building from a Table of Values

For a table, look for:

A clear vertex (a highest or lowest y-value in the middle).
Pairs of x-values equidistant from the vertex that share the same y-value.
Points where y = 0 for possible roots.

Example 6: Table Suggesting Vertex Form

A function f is quadratic. Some values are shown:

x   -1    0    1    2    3
f   -1    2    3    2   -1

Find a possible formula for f(x).

Show explanation

Notice:

Maximum value is 3 at x = 1.
Values are symmetric: f(0) = f(2) = 2, f(-1) = f(3) = -1.

That suggests a vertex at (1, 3). So try vertex form:

f(x) = a(x – 1)² + 3.

Use x = 0, f(0) = 2:

2 = a(0 – 1)² + 3

2 = a(1) + 3

a = -1.

So f(x) = -(x – 1)² + 3.

If you want standard form:

f(x) = -(x² – 2x + 1) + 3

= -x² + 2x – 1 + 3

= -x² + 2x + 2.

6. Full SAT-Style Examples: From Words/Graphs to Quadratics

These problems combine everything: building the model, choosing the right form, and answering the question.

Example 7: Rectangle Area Story

A rectangular garden has a fixed perimeter of 40 meters. The length is x meters, and the width is (20 – x) meters.
The area A(x) is quadratic in x.

(a) Write A(x) as a quadratic expression.
(b) What dimensions give the maximum area?

Show explanation

(a) Write A(x)
Area = length × width = x(20 – x).

So A(x) = 20x – x² or A(x) = -x² + 20x.

(b) Maximum area
This is a downward-opening parabola. Use vertex:

A(x) = -x² + 20x → a = -1, b = 20.

x = -b / (2a) = -20 / (2 · -1) = 10.

So the length is 10 meters. The width is 20 – x = 10 meters.

The maximum area occurs when the garden is a 10 by 10 square.

Example 8: Graph Interpretation

The graph of a quadratic function g(x) has x-intercepts at -2 and 6 and a maximum value of 8.
Which of the following could be the formula for g(x)?

(You won’t see the choices here, but we’ll build the general form.)

Show explanation

Roots at x = -2 and x = 6 → factored core:

g(x) = a(x + 2)(x – 6).

Since there is a maximum value (and we know it’s 8), the parabola opens downward → a < 0.
Also, the vertex is halfway between -2 and 6:

x = (-2 + 6) / 2 = 2.

Plug x = 2 into the factored expression to relate a and the max:

g(2) = a(2 + 2)(2 – 6) = a(4)(-4) = -16a.

But g(2) = 8 (the maximum).

So -16a = 8 → a = -1/2.

Final model:

g(x) = -\frac{1}{2}(x + 2)(x – 6).

In standard form, you could expand if answer choices are not factored.

Example 9: Data Fitting a Quadratic

A quadratic function h(x) passes through the points (0, 4), (2, 0), and (4, 4). Find a formula for h(x).

Show explanation

Notice:

Points (0, 4) and (4, 4) are symmetric around x = 2.
Point (2, 0) is in the middle and has the lowest y-value.

That suggests vertex at (2, 0). Also, (0, 4) is a point on the graph.

Start with vertex form:

h(x) = a(x – 2)² + 0 = a(x – 2)².

Use point (0, 4):

4 = a(0 – 2)² = a(4)

a = 1.

So h(x) = (x – 2)².

Check with (4, 4): (4 – 2)² = 4 → matches.

7. Common Traps & a Simple Practice Plan

Common Traps

Forgetting what x represents. Always label your variable in your head:
“x = time in seconds”, “p = price in dollars”, etc.
Mixing up vertex vs roots. Maximum/minimum → vertex. Starting and ending at ground level → roots.
Not expanding fully when needed. If the answer choices are in standard form,
expand your vertex or factored form carefully.
Ignoring units. The question might ask for time, height, or price. Don’t just give an x-value
without understanding what it means.

Practice Plan (30–40 minutes per session)

Do 3–4 word problems where you build the quadratic yourself
(projectiles, area, revenue). For each one, write your model and label what each part means.
Do 2–3 graph/table problems. Practice going from intercepts/vertex → equation,
or reading vertex/root information from the equation.
Add 2–3 mixed SAT questions where you must decide:
“Is this a vertex question, roots question, or value-at-x question?”
Keep a small modeling journal. Whenever you see a new type of story (e.g., profit, height, area),
save the structure so next time it feels familiar, not new.

Once you can quickly turn stories and graphs into clean quadratic models, a big chunk of “hard-looking”
SAT Math problems becomes much more manageable — and your path to 800 gets a lot smoother.