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Algebra I/Statistics/Modeling with Quadratics
Algebra I Regents in 22 days
Algebra I · Lesson 2

Modeling with Quadratics

Real situations that follow a curved path — and how to read them.

In 2016, Stephen Curry set an NBA record by making 402 three-pointers in a single season. Every one of those shots followed a curved path from his hand to the basket. That curve was not random — it was a parabola, shaped by gravity pulling the ball down while its forward momentum carried it forward. The height of the ball at any moment followed a quadratic equation. Quadratics are not just abstract algebra. They describe any situation where something grows, peaks, and then falls back down — or where two dimensions of measurement multiply together.

A quadratic function has the form

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

where a≠0a \neq 0. When you model a real situation with a quadratic, every piece of that equation means something. The constant cc is the starting value — the height of the ball when it left the player's hand, or the area of a garden before any additions are made. The vertex is the turning point — the maximum height of the ball, or the dimensions that produce the largest possible area. The zeros are where the function equals zero — where the ball hits the ground, or where a profit calculation breaks even.

The most common real-world quadratic is the projectile motion model. When an object is launched straight up, or at an angle, its height in feet after tt seconds follows:

h(t)=−16t2+v0t+h0h(t) = -16t^2 + v_0 t + h_0

The −16-16 comes from gravity pulling objects toward Earth at KaTeX can only parse string typed expression — the negative sign means height is decreasing due to gravity, and the 1616 is half of 3232, which appears because of how distance relates to acceleration. You do not need to derive that — just know that on the Regents, −16t2-16t^2 is the gravity term when height is in feet. The value v0v_0 is the initial upward velocity in feet per second. The value h0h_0 is the starting height in feet.

When a problem asks for the maximum height, you need the vertex. The tt-coordinate of the vertex gives the time when the object reaches its peak. Plug that time back into the equation and you get the maximum height.

A ball launched from the ground
h(t)=−16t2+64t+0h(t) = -16t^2 + 64t + 0
A ball is launched upward at 64 ft/s from ground level. Starting height is 0, so the constant term is 0.
t=−642(−16)t = -\frac{64}{2(-16)}
The vertex formula gives the time of maximum height. The t-coordinate of the vertex is t = -b divided by 2a.
t=−64−32=2t = -\frac{64}{-32} = 2
The ball reaches its peak at t = 2 seconds.
h(2)=−16(2)2+64(2)h(2) = -16(2)^2 + 64(2)
Plug t = 2 back in to find the actual height at that moment.
h(2)=−16(4)+128=−64+128=64 fth(2) = -16(4) + 128 = -64 + 128 = 64 \text{ ft}
The maximum height is 64 feet.

When a problem asks when the object hits the ground, you need the zeros. Set h(t)=0h(t) = 0 and solve. You can factor, complete the square, or use the quadratic formula — whichever works cleanest for the numbers in front of you.

When does the ball land?
−16t2+64t=0-16t^2 + 64t = 0
Set h(t) = 0. The ball is at height 0 when it starts and when it lands.
−16t(t−4)=0-16t(t - 4) = 0
Factor out -16t. Both factors share that.
t=0ort=4t = 0 \quad \text{or} \quad t = 4
t = 0 is when the ball was launched. t = 4 is when it hits the ground — that is the answer we want.
t=4 seconds✓t = 4 \text{ seconds} \checkmark
The ball lands 4 seconds after launch.

Quadratics also appear in area problems. If you have a fixed amount of fencing and want to enclose the largest possible rectangular area, the relationship between side length and total area turns out to be quadratic. The vertex of that quadratic gives you the dimensions that maximize the area.

Say a farmer has 80 feet of fencing and wants to enclose a rectangle against a barn wall, so she only needs to fence three sides. Let xx be the length of the side perpendicular to the barn. Then the side parallel to the barn has length 80−2x80 - 2x, and the total area is:

A(x)=x(80−2x)=−2x2+80xA(x) = x(80 - 2x) = -2x^2 + 80x

This is a quadratic with a negative leading coefficient, so its parabola opens downward and its vertex is the maximum.

Maximizing the fenced area
A(x)=−2x2+80xA(x) = -2x^2 + 80x
This is the area function. The variable x is the length of each side going away from the barn.
x=−802(−2)=−80−4=20x = -\frac{80}{2(-2)} = -\frac{80}{-4} = 20
Vertex formula gives the x that produces maximum area.
A(20)=−2(20)2+80(20)A(20) = -2(20)^2 + 80(20)
Plug x = 20 back in to find the maximum area.
A(20)=−2(400)+1600=−800+1600=800 ft2A(20) = -2(400) + 1600 = -800 + 1600 = 800 \text{ ft}^2
The maximum enclosed area is 800 square feet when the perpendicular sides are each 20 feet.
800 ft2✓800 \text{ ft}^2 \checkmark
The side parallel to the barn would be 80 - 2(20) = 40 feet long.

Here is what a projectile path looks like — the same ball from the first example, with its peak and landing point visible.

Interactive graph — scroll to zoom, drag to pan
Practice Questions
h(t)=−16t2+48t+0h(t) = -16t^2 + 48t + 0
A(x)=−3x2+60xA(x) = -3x^2 + 60x
−16t2+32t+48=0-16t^2 + 32t + 48 = 0
Regents Corner

On Part II and Part III of the Algebra I Regents, modeling problems ask you to interpret key features in context, not just calculate them. Writing "the vertex is (2, 64)" earns partial credit. Writing "the ball reaches its maximum height of 64 feet after 2 seconds" earns full credit. The numbers are the same — the interpretation is what the rubric checks for. Always attach units and always explain what the coordinate means in the situation.

A student solves a projectile problem, finds t = -1 and t = 3, and circles both as the answer. On the Regents, circling both without rejecting the negative value costs you the point. Time cannot be negative in a launch problem. You must state that t = -1 has no meaning in this context and that the answer is t = 3 seconds. One sentence of explanation is all it takes.
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Key Features of Quadratic Functions
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Square Root Functions