Parabolas, vertex, axis of symmetry, zeros, factoring, square roots, completing the square, the quadratic formula, the discriminant, and quadratic models — the first curved function family in Algebra I
This chapter introduces one of the most important function families in all of algebra: the quadratic function. Until now, every function you have studied has been linear — its graph is a straight line, its rate of change is constant, and its equation takes the form y = mx + b. Quadratics are fundamentally different. Their graphs curve. Their rates of change are not constant. They can model situations that linear functions cannot: the arc of a thrown ball, the area of a rectangle whose dimensions both change, the relationship between the side of a square and its area, and the revenue of a business whose sales respond to price.
This chapter is not only about solving equations. It is about understanding a function family. You will learn what makes a function quadratic, how to read and build its graph, how to find its key features, and how to solve quadratic equations using multiple methods. You will also learn to recognize when a quadratic model fits a real-world situation and how to interpret what the solution means in context.
Every method you learn in this chapter — factoring, square roots, completing the square as an idea, and the quadratic formula — is a tool. Each tool fits certain situations better than others. Part of becoming skilled at quadratics is learning which tool to reach for and why.
A soccer player kicks a ball from ground level. The height of the ball in feet, h, after t seconds is modeled by the equation h = −16t² + 48t. How high does the ball travel? When does it reach its highest point? When does it land? At what times is it exactly 32 feet high? None of these questions can be answered with the linear tools from earlier chapters. By the end of this chapter, you will be able to answer every one of them.
Quadratic functions appear throughout mathematics, science, engineering, and everyday life. The path of any projectile under gravity follows a quadratic curve. The area of any square or rectangle with variable dimensions involves a quadratic expression. The profit curve of a business, the shape of a satellite dish, the behavior of certain physical forces — all of these involve quadratic relationships.
Beyond applications, quadratic functions are the first major function family beyond linear, and they are the entry point to all of the higher mathematics that follows. Polynomial functions, complex numbers, conic sections, and much of calculus build directly on what you learn here. Understanding the parabola now means you are prepared for every curve that comes after it.
In earlier chapters, you studied linear functions. A linear function grows or decreases at a constant rate. Add the same amount to the input, and the output changes by a fixed amount every time. The graph of a linear function is always a straight line.
A quadratic function behaves differently. The rate of change is not constant — it speeds up or slows down depending on where you are on the graph. The graph is not a straight line; it curves. What makes a function quadratic is the presence of a squared variable as the highest-power term. The general form of a quadratic function is:
The coefficient a is called the leading coefficient. The term ax² is the quadratic term. The term bx is the linear term. The number c is the constant term.
The crucial requirement is that a ≠ 0. If a = 0, the quadratic term disappears and the function becomes linear. A quadratic function must have a nonzero squared term.
To see why the squared term changes everything, compare a linear and a quadratic function side by side.
| x | f(x) = 2x (linear) | Change | g(x) = x² (quadratic) | Change |
|---|---|---|---|---|
| 0 | 0 | — | 0 | — |
| 1 | 2 | +2 | 1 | +1 |
| 2 | 4 | +2 | 4 | +3 |
| 3 | 6 | +2 | 9 | +5 |
| 4 | 8 | +2 | 16 | +7 |
The linear function increases by exactly 2 each step. The quadratic function increases by more and more each step. This growing rate of change is what makes the graph curve rather than remain straight.
| Coefficient | Controls | Detail |
|---|---|---|
a | Direction and width of parabola | a > 0 opens upward; a < 0 opens downward; larger |a| = narrower parabola |
b | Horizontal position of vertex | Works with a to determine x = −b/(2a) |
c | y-intercept | When x = 0, f(0) = c; parabola always passes through (0, c) |
Do not confuse a function that contains x² with one where x² is the leading term. In x³ + x², the degree is 3, not 2. The highest power determines the function type.
Every quadratic function has the same basic graph shape: a smooth, symmetric curve called a parabola. You have likely seen this shape before — it looks like the path of a ball thrown through the air, or the inside of a bowl, or the outline of a satellite dish. A parabola has several important features: it is symmetric about a central vertical line, it has a single turning point called the vertex, and it opens either upward or downward.
If a > 0, the parabola opens upward. The vertex is at the bottom — a minimum.
If a < 0, the parabola opens downward. The vertex is at the top — a maximum.
This matters for applications: if a quadratic models height of a thrown object, the vertex gives the maximum height. If it models profit, the vertex gives maximum profit (when a < 0) or minimum cost (when a > 0).
The value of |a| controls how wide or narrow the parabola is. A large |a| pulls the parabola tighter; a small |a| close to zero spreads it open. For example, f(x) = 3x² is narrower than f(x) = x², which is narrower than f(x) = (1/4)x².
Draw two parabolas on the same coordinate plane. The first is f(x) = x², a standard upward parabola with vertex at the origin. The second is f(x) = −x², a downward parabola with vertex at the origin. Label the vertex (0, 0) for each, mark the axis of symmetry x = 0 with a dashed line, and indicate the direction each opens. Mark the y-intercept (0, 0) for both. Add arrows at the ends of each curve to show the parabola continues.
The vertex is the most important single point on a parabola. It represents the function's maximum or minimum value — the highest point a ball reaches, the minimum cost of production, the maximum area that can be enclosed with a fixed perimeter. For any quadratic function f(x) = ax² + bx + c, the x-coordinate of the vertex is:
Once you have the x-coordinate, substitute it into the function to find the y-coordinate: y = f(−b/(2a)). The axis of symmetry is the vertical line x = −b/(2a) — it passes directly through the vertex and divides the parabola into two mirror-image halves.
The axis of symmetry is x = −b/(2a), not b/(2a). The negative sign is always part of the formula. If b = −6, then −b/(2a) = −(−6)/(2a) = 6/(2a), which is positive. Write out the sign substitution explicitly to avoid errors.
x = −(−6) / (2 · 1) = 6 / 2 = 3
f(3) = (3)² − 6(3) + 5 = 9 − 18 + 5 = −4
Vertex: (3, −4). Axis of symmetry: x = 3.
x = −8 / (2 · (−2)) = −8 / (−4) = 2
f(2) = −2(2)² + 8(2) − 3 = −8 + 16 − 3 = 5
Vertex: (2, 5).
Before using shortcuts to graph a quadratic, it is valuable to build the graph directly from a table of values. This reveals the shape and symmetry of the parabola and reinforces why graphing by key features works. Choose x-values that include the vertex and extend to both sides of the axis of symmetry. Plot the resulting ordered pairs, then connect them with a smooth curve — never with straight line segments. A parabola is a continuously curving shape, not a polygon.
x = 0: 0 − 0 + 3 = 3 → (0, 3)
x = 1: 1 − 4 + 3 = 0 → (1, 0)
x = 2: 4 − 8 + 3 = −1 → (2, −1) ← vertex
x = 3: 9 − 12 + 3 = 0 → (3, 0)
x = 4: 16 − 16 + 3 = 3 → (4, 3)
Notice: the values at x = 1 and x = 3 are both 0. The values at x = 0 and x = 4 are both 3. The table is symmetric about x = 2.
Draw a coordinate plane. Plot the five points (0, 3), (1, 0), (2, −1), (3, 0), (4, 3). Draw a dashed vertical line at x = 2 labeled "axis of symmetry." Connect the points with a smooth upward-opening U-shaped curve. Label the vertex (2, −1), the two x-intercepts (1, 0) and (3, 0), and the y-intercept (0, 3). Add arrows at both ends of the parabola to show it continues.
Always connect plotted points with a smooth curve, not jagged line segments. The parabola has no sharp corners. If the curve looks angular in your sketch, add more plotted points between the ones you have.
When you need to graph a quadratic quickly and accurately, you can use key features of the standard form rather than building a full table. The strategy below produces a well-labeled, accurate graph efficiently.
1. Identify a, b, and c. Determine direction (a > 0 opens upward; a < 0 opens downward).
2. Find the axis of symmetry: x = −b/(2a).
3. Find the vertex by substituting the x-coordinate into f(x).
4. Find the y-intercept: let x = 0, giving (0, c).
5. Use symmetry to find the mirror of the y-intercept across the axis.
6. Find x-intercepts if they exist (set f(x) = 0 and solve).
7. Draw a smooth parabola through the key points.
Axis of symmetry: x = −2 / (2·(−1)) = 1
Vertex: f(1) = −1 + 2 + 3 = 4 → (1, 4) [maximum]
y-intercept: f(0) = 3 → (0, 3)
Mirror of y-intercept across x = 1: x = 2 → (2, 3)
For x-intercepts, set f(x) = 0:−x² + 2x + 3 = 0 → x² − 2x − 3 = 0
(x − 3)(x + 1) = 0 → x = 3 or x = −1
X-intercepts: (3, 0) and (−1, 0).
The x-intercepts of a quadratic function are the points where the parabola crosses the x-axis. At these points, the output of the function equals zero. A value of x that makes f(x) = 0 is called a zero of the function. When zeros are real numbers, they correspond exactly to the x-intercepts. The word root means the same thing in the context of equations.
These three terms — x-intercept, zero, and root — describe the same values from three perspectives: geometric, functional, and algebraic.
| Discriminant | Number of x-Intercepts | What it Looks Like |
|---|---|---|
| D > 0 | Two distinct x-intercepts | Parabola crosses the x-axis at two separate points |
| D = 0 | One x-intercept (at the vertex) | Parabola touches the x-axis at exactly one point |
| D < 0 | No x-intercepts | Parabola does not touch the x-axis at all |
The zeros of a quadratic function are symmetric about the axis of symmetry. If the zeros are at x = r and x = s, then the axis of symmetry passes through their midpoint: x = (r + s)/2.
Axis of symmetry: x = (−2 + 6) / 2 = 4 / 2 = 2
To solve a quadratic equation by factoring, you need the equation in standard form with everything on one side and zero on the other. The method relies on the Zero Product Property: if two factors multiply to zero, at least one of them must equal zero.
If A · B = 0, then A = 0 or B = 0 (or both).
This property is the entire reason factoring works as a solution method. It converts one equation into two simpler equations, each easily solved.
Critical: The product must equal zero. If the product equals any other number, this property does not apply. Move all terms to one side first.
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
x − 2 = 0 or x − 3 = 0
x = 2 or x = 3
x² + 3x − 10 = 0
(x + 5)(x − 2) = 0
x = −5 or x = 2
2x² − 6x − x + 3 = 0
2x(x − 3) − 1(x − 3) = 0
(2x − 1)(x − 3) = 0
x = 1/2 or x = 3
The Zero Product Property only works when the product equals zero. If you have (x + 3)(x − 2) = 6, you cannot set each factor equal to 6. You must first expand: x² + x − 6 = 6, then rearrange to x² + x − 12 = 0, then factor and solve.
Some quadratic equations have no x term — only x² and a constant. Equations of the form x² = k or (x + d)² = k can be solved directly by taking square roots, which is faster than factoring. The key principle is the square root property.
If x² = k and k ≥ 0, then x = √k or x = −√k.
Written compactly: x = ±√k
The ± symbol is essential. Both the positive and negative square roots are valid solutions. Omitting the negative root is one of the most common errors in solving quadratics.
x = ±√49 = ±73x² = 27
x² = 9
x = ±3
x − 3 = ±√16 = ±4
Case 1: x − 3 = 4 → x = 7
Case 2: x − 3 = −4 → x = −1
x² = −9There is no real number whose square is −9. Squaring any real number gives a result ≥ 0.Completing the square is an algebraic technique for rewriting a quadratic expression so that it contains a perfect square trinomial — a trinomial that factors into (x + d)² for some value d. At the Algebra I level, you need to understand what completing the square means and why it works, because it is the foundation of the quadratic formula and of vertex form. A full mastery of completing the square as a solution procedure is a central topic of Algebra II Chapter 3.
A perfect square trinomial is a trinomial of the form x² + 2dx + d² = (x + d)². The pattern: the constant term is always the square of half the coefficient of x. For example, in x² + 6x + 9: half of 6 is 3, and 3² = 9. Adding 9 completes the square.
| Expression | Completing Value (half of b)² | Perfect Square Form |
|---|---|---|
x² + 6x | (6/2)² = 9 | (x + 3)² |
x² − 10x | (−10/2)² = 25 | (x − 5)² |
x² + 4x | (4/2)² = 4 | (x + 2)² |
f(x) = (x² − 6x) + 5
= (x² − 6x + 9) − 9 + 5
= (x − 3)² − 4
Vertex form: f(x) = (x − 3)² − 4. Vertex: (3, −4).
Factoring works beautifully when the quadratic factors nicely over the integers. But many quadratic equations — in fact, most real-world quadratics — do not factor cleanly. The quadratic formula solves any quadratic equation, regardless of whether it factors.
This formula applies to any quadratic equation written in standard form ax² + bx + c = 0, where a ≠ 0. The ± sign means the formula produces two values:
x₁ = (−b + √(b² − 4ac)) / (2a) x₂ = (−b − √(b² − 4ac)) / (2a)
The formula is derived by applying completing the square to the general standard form — a process you will study fully in Algebra II. The formula is not magic; it is compressed algebra.
1. Write the equation in standard form: ax² + bx + c = 0.
2. Identify a, b, and c.
3. Compute the discriminant b² − 4ac first.
4. Substitute into the formula.
5. Simplify the square root.
6. Compute both solutions using + and −.
x = [ −(−5) ± √((−5)² − 4·1·4) ] / (2·1)
= [ 5 ± √(25 − 16) ] / 2
= [ 5 ± √9 ] / 2
= [ 5 ± 3 ] / 2
x = (5 + 3)/2 = 4 or x = (5 − 3)/2 = 1
x = [ −4 ± √(16 − 4) ] / 2
= [ −4 ± √12 ] / 2
= [ −4 ± 2√3 ] / 2
= −2 ± √3
Solutions: x = −2 + √3 ≈ −0.27 or x = −2 − √3 ≈ −3.73
3x² − 2x − 8 = 0
a = 3, b = −2, c = −8.x = [ 2 ± √(4 + 96) ] / 6
= [ 2 ± √100 ] / 6
= [ 2 ± 10 ] / 6
x = 12/6 = 2 or x = −8/6 = −4/3
Compute b² − 4ac first. Then simplify the square root. Then calculate both values separately. Working through the formula in strict order prevents the most common arithmetic mistakes.
Before solving a quadratic, you can quickly determine how many real solutions to expect by computing the discriminant — the expression under the radical in the quadratic formula:
| If D… | Number of Real Solutions | Graph Description |
|---|---|---|
| D > 0 | Two distinct real solutions | Parabola crosses x-axis at two points |
| D = 0 | One real solution (repeated root) | Parabola touches x-axis at exactly one point (vertex) |
| D < 0 | No real solutions | Parabola does not touch the x-axis |
When D < 0, the equation has no real solutions because no real number has a negative square. In Algebra II Chapter 4, you will learn about complex numbers, which allow equations like x² = −9 to have solutions.
Many real-world situations lead naturally to quadratic equations or require interpreting a quadratic function. The key is recognizing when a situation involves a squared quantity, a product of two variable expressions, or a known physics formula. Use the Read → Represent → Solve → Check → Explain method throughout.
The height of any object launched upward or dropped near Earth's surface follows a quadratic model. In feet and seconds, the general formula is:
where h₀ is the initial height in feet and v₀ is the initial upward velocity in feet per second. In metric units (meters and seconds), the coefficient −16 is replaced by −4.9.
t = −64 / (2 · (−16)) = −64 / (−32) = 2 seconds
h(2) = −16(4) + 64(2) = −64 + 128 = 64 feet
−16t² + 64t = 0
−16t(t − 4) = 0
t = 0 or t = 4
w² + 3w = 40
w² + 3w − 40 = 0
(w + 8)(w − 5) = 0
w = −8 or w = 5
Discard w = −8 (width cannot be negative). So w = 5 meters, length = 8 meters.
p = −200 / (2 · (−10)) = −200 / (−20) = 10 dollars
R(10) = 200(10) − 10(100) = 2000 − 1000 = 1000 dollars
Correct: x² = 36 → x = 6 or x = −6. Both (6)² and (−6)² equal 36. Always write ±.
Correct: Expand, rearrange to standard form (x² + x − 6 = 12 → x² + x − 18 = 0), then factor or use the quadratic formula.
Correct: x = −(−8)/(2·1) = 8/2 = 4. The formula is −b/(2a), not b/(2a). The negative sign is always part of the formula.
Correct: Move all terms left first. x² + 5x − 3 = 0. Now c = −3, not +3.
Correct: If b = −5, then −b = −(−5) = +5. Write it out explicitly: −(−5) = 5.
Correct: Always simplify. √12 = √(4·3) = 2√3. Simplified radical form is standard unless a decimal approximation is specifically requested.
Correct: In pure equation solving, both solutions are valid. Only discard a negative value when the context makes it meaningless (e.g., a negative width, a negative time before launch).
Correct: b² − 4ac is a single numerical computation. Calculate b², calculate 4ac, then subtract. Do not try to factor or simplify prematurely.
Correct: A parabola can have zero, one, or two x-intercepts. Always check the discriminant. If D < 0, the parabola does not touch the x-axis and the equation has no real solutions.
Correct: They are the same. A zero of f is a value where f(x) = 0. A root is a solution to the equation ax² + bx + c = 0. An x-intercept is where the graph crosses the x-axis. All three describe the same values.
In this chapter, you have studied the quadratic function as one of the most important mathematical structures in Algebra I. You can identify a quadratic function, describe its graph, find the vertex and axis of symmetry, build a table of values, and graph using key features. You can solve quadratic equations by four methods: factoring, square roots, completing the square as an idea, and the quadratic formula. You can use the discriminant to determine the number of solutions and apply quadratic models to area, projectile, and revenue problems.
Vertex form and transformations: Algebra II Chapter 3 introduces f(x) = a(x − h)² + k as a primary graphing form. You will study how changing a, h, and k shifts, stretches, and reflects the parabola — the same transformation language applied to every function family.
Completing the square as a procedure: In Algebra II, you will complete the square as a full algebraic technique to convert between forms and solve equations, not just as an idea.
Deriving the quadratic formula: Algebra II Chapter 3 walks through the full derivation by completing the square on ax² + bx + c = 0, showing you exactly where the formula comes from.
Complex solutions: When D < 0, Algebra II Chapter 4 provides the tools — complex numbers — to work with solutions that are not real numbers.
Optimization: More complex maximum and minimum problems drawn from economics, engineering, and science.
| Course | How Quadratics Appear |
|---|---|
| Geometry | Area and perimeter formulas; the Pythagorean theorem produces quadratic equations; conic sections are defined by quadratic equations |
| Physics | Projectile motion; kinetic energy (KE = ½mv²); gravitational potential energy; wave functions |
| Finance | Revenue, cost, and profit models; quadratic growth in certain investment scenarios |
| Computer Science | Quadratic-time algorithms (O(n²)); simulation of physical systems; collision detection |
| Precalculus and Calculus | Vertex form, completing the square, and factoring are used throughout; limits of quadratics; optimization problems in calculus |
Identify each as quadratic or not. For each quadratic, state a, b, c, and the direction of opening.
f(x) = 3x² − x + 7g(x) = 5x − 9h(x) = −x²k(x) = x³ + 2x² − 1m(x) = −2x² + 4xn(x) = 0·x² + 4x + 1Find the vertex and axis of symmetry. State whether the vertex is a minimum or maximum.
f(x) = x² + 4x − 5f(x) = −3x² + 6x + 1f(x) = 2x² − 8x + 9f(x) = x² − 10f(x) = −x² + 2x + 5f(x) = 4x² + 16x + 7For each function, identify the vertex, axis of symmetry, y-intercept, direction, and x-intercepts (if they exist). Sketch the graph.
f(x) = x² − 2x − 8f(x) = −x² + 4xf(x) = x² + 6x + 9f(x) = 2x² − 4x − 6x² + 7x + 12 = 0x² − 9x + 20 = 0x² − x − 30 = 02x² + 5x − 3 = 03x² − 12x = 0x² = 4x + 216x² + x − 2 = 0x² − 16 = 0x² = 100x² − 18 = 0(x + 5)² = 494x² = 36(2x − 1)² = 25x² + 16 = 0Rewrite in vertex form by completing the square. Identify the vertex.
f(x) = x² + 8x + 12f(x) = x² − 10x + 21f(x) = x² + 2x − 5f(x) = x² − 4x + 1Solve each equation using the quadratic formula. Simplify all answers.
x² + 5x + 6 = 0x² − 3x − 1 = 02x² + 3x − 2 = 0x² − 6x + 9 = 03x² + x + 2 = 02x² = 5x + 3Find the discriminant and state the number of real solutions.
x² − 4x + 4 = 0x² + 2x + 5 = 03x² − 7x + 2 = 04x² + x + 1 = 05x² − 20 = 0x² − 6x + 9 = 0Use the Read → Represent → Solve → Check → Explain method for each problem. Show all steps. Define variables clearly.
Each problem shows a student's work. Find the error. Explain what went wrong. Write the correct solution.
What did the student omit? What is the complete correct answer?
What property was misapplied? Describe the error and show the correct approach.
What sign error occurred in the formula? Find the correct vertex.
What setup error was made? What should a, b, and c be?
Is the calculation correct? Is the conclusion correct? Explain the error in the conclusion.
f(x) = −4x² + 7x − 2 a quadratic function? State a, b, c and the direction of opening.f(x) = x² − 2x − 8.f(x) = −2x² + 4x − 1 is a minimum or maximum.f(x) = x² + 2x − 3. Label the vertex, axis of symmetry, y-intercept, and x-intercepts.x² − 8x + 15 = 0(x − 2)² = 36x² − 4x + 1 = 02x² + x + 5 = 0 and state the number of real solutions.x² − 6x + 10 = 0 has two real solutions. Use the discriminant to confirm or correct this claim.h(t) = −16t² + 80t. (a) Find the maximum height. (b) Find when the ball returns to the ground.The quadratic formula contains a square root: √(b² − 4ac). Chapter 13 deepens your understanding of square roots, cube roots, and radical expressions. After Chapter 13, you will simplify radical results from the quadratic formula more fluently — for example, expressing √12 as 2√3 or working with expressions like √48 and 3√5 with confidence. You will also connect radicals to the Pythagorean theorem and coordinate geometry, and solve simple radical equations. All of this prepares you for the more advanced treatment of quadratic solutions in Algebra II.