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PrecalculusStewart Ch. 3.1

Quadratic Functions

Master vertex form, completing the square, graphing parabolas, and real-world applications. Step-by-step for high school and early college students.

Standard & vertex formVertex formulaCompleting the squareMax and min valuesProjectile motion

On This Page

  1. 1. Standard Form
  2. 2. Vertex Form
  3. 3. Finding the Vertex
  4. 4. The Sign Trap
  5. 5. Axis of Symmetry
  6. 6. Maximum and Minimum Values
  7. 7. Completing the Square
  8. 8. Converting Between Forms
  9. 9. Graphing Parabolas
  10. 10. Real-World Applications
  11. 11. FAQ

1. Standard Form

The standard form (also called general form) of a quadratic function is:

f(x) = ax² + bx + c

where a ≠ 0

The three constants a, b, and c each play a distinct role:

  • a
    Leading coefficient a

    Controls the direction and width of the parabola. If a is positive, the parabola opens upward. If a is negative, it opens downward. The larger the absolute value of a, the narrower the parabola; the smaller, the wider.

  • b
    Middle coefficient b

    Together with a, determines the horizontal position of the vertex via the formula x = −b / (2a). By itself it does not have a simple geometric meaning, but the ratio −b/(2a) is one of the most important formulas in all of precalculus.

  • c
    Constant term c

    The y-intercept of the parabola. When you substitute x = 0 into standard form, the ax² and bx terms both vanish and you get f(0) = c. So the parabola always crosses the y-axis at the point (0, c).

Examples in standard form

  • f(x) = 3x² − 5x + 2   →   a = 3, b = −5, c = 2
  • g(x) = −x² + 4x   →   a = −1, b = 4, c = 0
  • h(x) = x² − 9   →   a = 1, b = 0, c = −9
  • p(x) = −2x² − 6x − 1   →   a = −2, b = −6, c = −1

2. Vertex Form

Vertex form reveals the turning point of the parabola immediately without any calculation:

f(x) = a(x − h)² + k

Vertex is the point (h, k). Axis of symmetry is x = h.

What h and k mean geometrically

h — horizontal shift

h shifts the basic parabola y = ax² left or right. Positive h moves the vertex to the right of the origin; negative h moves it to the left. The axis of symmetry is the vertical line x = h.

k — vertical shift

k shifts the parabola up or down. Positive k raises the vertex above the x-axis; negative k lowers it below. The minimum or maximum value of f is exactly k, and it occurs at x = h.

Reading vertex form at a glance

EquationahkVertexOpens
2(x − 3)² + 5235(3, 5)Up
−(x + 4)² + 1−1−41(−4, 1)Down
½(x − 1)² − 6½1−6(1, −6)Up
−3(x − 0)² + 8−308(0, 8)Down

3. Finding the Vertex

There are two methods, depending on which form you start with.

Method A: From Standard Form

Use the vertex formula. Given f(x) = ax² + bx + c:

x‑coordinate of vertex = −b / (2a)

Substitute back into f(x) to get the y‑coordinate.

Worked Example — Find the vertex of f(x) = 2x² − 8x + 3

  1. 1Identify: a = 2, b = −8, c = 3.
  2. 2x = −b / (2a) = −(−8) / (2 × 2) = 8 / 4 = 2
  3. 3f(2) = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5
  4. 4Vertex is (2, −5).

Method B: From Vertex Form

If the equation is already in vertex form f(x) = a(x − h)² + k, the vertex is simply the point (h, k). No calculation required — just read h and k directly from the equation. (But watch out for the sign trap explained in the next section.)

Quick reads:

  • f(x) = 4(x − 7)² + 2  →  vertex (7, 2)
  • f(x) = −(x + 3)² − 1  →  rewrite as −(x − (−3))² − 1  →  vertex (−3, −1)
  • f(x) = 3x² + 5  →  rewrite as 3(x − 0)² + 5  →  vertex (0, 5)

4. The Sign Trap

Common Mistake — Read This Carefully

Vertex form is f(x) = a(x − h)² + k. The formula uses a minus sign. This means the vertex x-coordinate is h, not −h. Students who forget the minus sign flip the sign of h and get the wrong vertex.

Correct Readings

  • (x − 3)²  →  vertex at x = 3
  • (x − 5)²  →  vertex at x = 5
  • (x + 2)² = (x − (−2))²  →  vertex at x = −2
  • (x + 7)² = (x − (−7))²  →  vertex at x = −7

Wrong Readings (Do Not Do This)

  • (x − 3)²  →  vertex at x = −3 ✗
  • (x − 5)²  →  vertex at x = −5 ✗
  • (x + 2)²  →  vertex at x = 2 ✗
  • (x + 7)²  →  vertex at x = 7 ✗

Foolproof check

Ask yourself: “What value of x makes the quantity inside the parentheses equal to zero?” That value is always h, the x-coordinate of the vertex. For (x − 3)², set x − 3 = 0 and solve: x = 3. For (x + 2)², set x + 2 = 0 and solve: x = −2.

5. Axis of Symmetry

Every parabola has exactly one axis of symmetry: a vertical line that divides the parabola into two identical mirror-image halves. The axis of symmetry always passes through the vertex.

Axis of symmetry:   x = h = −b / (2a)

The axis of symmetry is extremely useful for graphing. Once you know the vertex and have plotted any other point on the parabola, that point has a mirror image on the opposite side of the axis at the same height. This halves the work of sketching.

Worked Example

Find the axis of symmetry for f(x) = −3x² + 6x − 1.

  1. a = −3, b = 6
  2. x = −b / (2a) = −6 / (2 × (−3)) = −6 / (−6) = 1
  3. Axis of symmetry: x = 1

If f(0) = −1 (the y-intercept), then by symmetry the point (2, −1) is also on the parabola, because x = 0 and x = 2 are both 1 unit away from the axis x = 1.

6. Maximum and Minimum Values

Because a parabola has exactly one turning point (the vertex), the y-coordinate of the vertex is either the global maximum or global minimum of the function.

↟

a > 0 — Opens Up

  • Parabola opens upward
  • Vertex is the lowest point
  • f has a minimum value
  • Minimum value = k (the y-coordinate of the vertex)
  • Occurs at x = h
  • Range is [k, +∞)

↡

a < 0 — Opens Down

  • Parabola opens downward
  • Vertex is the highest point
  • f has a maximum value
  • Maximum value = k (the y-coordinate of the vertex)
  • Occurs at x = h
  • Range is (−∞, k]

Worked Example

Find the maximum or minimum of f(x) = −4x² + 16x − 3 and state the range.

  1. a = −4 < 0, so the parabola opens down and the vertex is a maximum.
  2. x = −b / (2a) = −16 / (2 × (−4)) = −16 / (−8) = 2
  3. f(2) = −4(4) + 16(2) − 3 = −16 + 32 − 3 = 13
  4. Maximum value is 13, occurring at x = 2. Range is (−∞, 13].

7. Completing the Square

Completing the square is the algebraic technique that converts standard form into vertex form. Once mastered, it unlocks the vertex and axis of symmetry without memorizing any additional formulas. It also proves the quadratic formula. Stewart Precalculus Chapter 3.1 uses this method as the primary tool for analyzing quadratic functions.

General Procedure

Given f(x) = ax² + bx + c with a ≠ 0:

  1. Step 1: Factor a out of the first two terms: f(x) = a(x² + (b/a)x) + c
  2. Step 2: Compute the completing number: (b/(2a))²
  3. Step 3: Add and subtract that number inside the parentheses
  4. Step 4: Factor the perfect square trinomial: (x + b/(2a))²
  5. Step 5: Distribute a and combine constants

Step-by-Step Worked Example

Convert f(x) = 2x² − 12x + 7 to vertex form.

1

Write the function

f(x) = 2x² − 12x + 7

Identify a = 2, b = −12, c = 7.

2

Factor a from the x-terms

f(x) = 2(x² − 6x) + 7

Factor 2 out of the first two terms only. The +7 stays outside.

3

Find the completing number

(−6 ÷ 2)² = (−3)² = 9

Take half the coefficient of x, then square it.

4

Add and subtract inside the parentheses

f(x) = 2(x² − 6x + 9 − 9) + 7

Adding 9 inside forces you to subtract 9 to keep the expression equal.

5

Rewrite the perfect square trinomial

f(x) = 2((x − 3)² − 9) + 7

x² − 6x + 9 = (x − 3)²

6

Distribute a and simplify

f(x) = 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11

Vertex is (3, −11). Since a = 2 > 0, the parabola opens up. Minimum value is −11.

Final Answer

f(x) = 2(x − 3)² − 11

Vertex: (3, −11). Opens up (a = 2 > 0). Minimum value: −11. Axis of symmetry: x = 3.

When a ≠ 1: The Critical Step

The most common error occurs when a is not 1 and students forget to factor it out before computing the completing number. Here is why the factor matters:

For f(x) = 3x² + 18x + 5, factor 3 out of the x-terms first:

f(x) = 3(x² + 6x) + 5

Now complete the square on x² + 6x. Half of 6 is 3, squared is 9. Add and subtract 9 inside:

f(x) = 3(x² + 6x + 9 − 9) + 5 = 3((x + 3)² − 9) + 5

f(x) = 3(x + 3)² − 27 + 5 = 3(x + 3)² − 22

Vertex: (−3, −22). The −27 came from distributing the 3 through −9 × 3 = −27.

8. Converting Between Forms

FormEquationVertex x-coordinateBest Used For
Standard Formf(x) = ax² + bx + cx = −b / (2a)Reading a, b, c; finding y-intercept quickly
Vertex Formf(x) = a(x − h)² + kx = h (read directly)Reading vertex (h, k) and axis of symmetry immediately

Standard Form → Vertex Form

Use completing the square (see Section 7) or the direct vertex formula:

h = −b/(2a)

k = c − b²/(4a)

Then write: f(x) = a(x − h)² + k

Vertex Form → Standard Form

Expand and simplify by multiplying out (x − h)²:

f(x) = a(x − h)² + k

= a(x² − 2hx + h²) + k

= ax² − 2ahx + ah² + k

So b = −2ah and c = ah² + k.

Worked Example: Vertex Form to Standard Form

Expand f(x) = −2(x + 1)² + 5.

  1. Expand (x + 1)² = x² + 2x + 1
  2. Multiply by −2: −2x² − 4x − 2
  3. Add the +5: f(x) = −2x² − 4x + 3
  4. Standard form: f(x) = −2x² − 4x + 3

Verify: vertex from standard form is x = −(−4) / (2 × −2) = 4 / (−4) = −1. Then f(−1) = −2(1) − 4(−1) + 3 = −2 + 4 + 3 = 5. Vertex (−1, 5) matches vertex form. ✓

9. Graphing Parabolas

Graphing a quadratic function systematically requires identifying six key pieces of information. Work through them in the order below to build a complete and accurate sketch.

1

Find the vertex

Use x = −b/(2a) from standard form, or read (h, k) directly from vertex form. The vertex is the turning point of the parabola.

2

Determine direction of opening

If a > 0 the parabola opens upward (minimum at vertex). If a < 0 it opens downward (maximum at vertex). Larger |a| means a narrower parabola.

3

Find the y-intercept

Substitute x = 0. In standard form f(0) = c, so the y-intercept is always the constant term. Plot (0, c).

4

Find x-intercepts if they exist

Solve ax² + bx + c = 0 by factoring, completing the square, or the quadratic formula. If the discriminant b² − 4ac is negative, there are no real x-intercepts.

5

Draw the axis of symmetry

Draw the vertical dashed line x = h through the vertex. Use this line to reflect points from one side of the parabola to the other to create a symmetric sketch.

6

Plot additional points and connect

Choose x-values on one side of the vertex, evaluate f(x), plot those points, then mirror them across the axis of symmetry. Connect with a smooth U-shaped curve.

Worked Graphing Example

Graph f(x) = x² − 4x − 5.

Finding key features:

  • a, b, c: a = 1, b = −4, c = −5
  • Direction: a = 1 > 0, opens up
  • Vertex x: −(−4) / (2 × 1) = 2
  • Vertex y: f(2) = 4 − 8 − 5 = −9
  • Vertex: (2, −9)
  • Axis of symmetry: x = 2
  • y-intercept: f(0) = −5 → (0, −5)

Finding x-intercepts:

  • Set x² − 4x − 5 = 0
  • Factor: (x − 5)(x + 1) = 0
  • x = 5 or x = −1
  • x-intercepts: (5, 0) and (−1, 0)
  • Check symmetry: 5 and −1 average to (5 + (−1)) / 2 = 2. ✓ (equals the axis of symmetry)
Summary of points to plot: vertex (2, −9), y-intercept (0, −5) and its mirror image (4, −5) by symmetry, x-intercepts (−1, 0) and (5, 0). Connect with a smooth upward-opening U-curve.

10. Real-World Applications

Quadratic functions model a wide variety of real situations. The three most common in precalculus are projectile motion, revenue maximization, and area optimization.

Application 1: Projectile Motion

When a ball, rocket, or any object is launched upward (in feet, ignoring air resistance), its height at time t seconds is modeled by:

h(t) = −16t² + v₀t + h₀

v₀ is the initial velocity in ft/s; h₀ is the initial height in feet. The −16 comes from half of the gravitational acceleration −32 ft/s².

Worked Example — A ball is launched from the ground with initial velocity 64 ft/s. Find the maximum height and the time when it lands.

h(t) = −16t² + 64t

  1. Vertex x (time of maximum):t = −b / (2a) = −64 / (2 × (−16)) = −64 / (−32) = 2 seconds
  2. Maximum height:h(2) = −16(4) + 64(2) = −64 + 128 = 64 feet
  3. Time of landing: Set h(t) = 0: −16t² + 64t = 0 → −16t(t − 4) = 0 → t = 0 (launch) or t = 4 (landing)
  4. The ball reaches 64 feet at t = 2 s and lands at t = 4 s.

Application 2: Revenue Maximization

Revenue = price × quantity sold. When demand depends linearly on price, the revenue function becomes quadratic and the vertex gives the optimal price.

Worked Example — A concert venue sells 500 tickets at $30 each. For every $1 increase in price, 10 fewer tickets are sold. Find the price that maximizes revenue.

  1. Set up variables: Let x = number of $1 price increases. Price per ticket: 30 + x. Tickets sold: 500 − 10x.
  2. Revenue function:R(x) = (30 + x)(500 − 10x)
  3. Expand:R(x) = 15000 − 300x + 500x − 10x² = −10x² + 200x + 15000
  4. Vertex:x = −200 / (2 × −10) = −200 / (−20) = 10
  5. Optimal price: 30 + 10 = $40 per ticket. Tickets sold: 500 − 100 = 400.
  6. Maximum revenue: R(10) = −10(100) + 200(10) + 15000 = −1000 + 2000 + 15000 = $16,000
  7. Maximum revenue is $16,000 when tickets are priced at $40.

Application 3: Area Optimization

Many geometry problems ask for the maximum area given a fixed perimeter. These lead naturally to quadratic functions.

Worked Example — You have 80 ft of fencing to enclose a rectangular garden against a wall (only three sides need fencing). Find the dimensions that maximize area.

  1. Variables: Let w = width (the two sides perpendicular to the wall). The side parallel to the wall has length 80 − 2w.
  2. Area function:A(w) = w × (80 − 2w) = 80w − 2w²
  3. Rewrite in standard form:A(w) = −2w² + 80w (opens down, so vertex is a maximum)
  4. Vertex:w = −80 / (2 × −2) = −80 / (−4) = 20 ft
  5. Length parallel to wall: 80 − 2(20) = 40 ft
  6. Maximum area: 20 × 40 = 800 ft²
  7. Optimal dimensions: 20 ft by 40 ft, for a maximum area of 800 ft².

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Frequently Asked Questions

Worked answers to the questions students most often have about quadratic functions in precalculus.

What is a quadratic function and how is it different from linear?+

A quadratic function has the form f(x) = ax² + bx + c with a ≠ 0. The highest power of x is 2, which is what makes it quadratic. A linear function has highest power 1 and graphs as a straight line.

The key differences: (1) the graph of a quadratic is a parabola (curved, U-shaped or upside-down U), not a line; (2) a quadratic can have 0, 1, or 2 real zeros, whereas a non-horizontal line always has exactly 1; (3) a quadratic has a vertex (turning point), a linear function does not.

How do I find the vertex from standard form without completing the square?+

Use the vertex formula directly. Given f(x) = ax² + bx + c:

  1. Compute x = −b / (2a). This is the x-coordinate of the vertex.
  2. Substitute that x-value into f(x). The result is the y-coordinate.

Example: f(x) = −x² + 6x − 2. Here a = −1, b = 6. x = −6 / (2 × −1) = −6 / (−2) = 3. f(3) = −9 + 18 − 2 = 7. Vertex: (3, 7).

What does the discriminant tell me about the graph?+

The discriminant is D = b² − 4ac, the expression under the square root in the quadratic formula. It tells you how many real x-intercepts the parabola has:

  • D > 0:Two distinct real zeros → parabola crosses the x-axis at two points.
  • D = 0:Exactly one real zero (repeated) → vertex sits exactly on the x-axis.
  • D < 0:No real zeros → parabola does not touch the x-axis at all.
Why do we add AND subtract the same number when completing the square?+

Because we must not change the value of the expression. Adding and then subtracting the same number is equivalent to adding zero, so the function is unchanged.

Example: starting from x² − 6x, we need to add 9 to create the perfect square (x − 3)². But just writing x² − 6x + 9 would give a different function. So we also subtract 9:

x² − 6x + 9 − 9 = (x − 3)² − 9

The two expressions x² − 6x and (x − 3)² − 9 are identical — they produce the same output for every input x. We just rewrote the same function in a more useful form.

How do I find the range of a quadratic function?+

First find the vertex (h, k). Then use the sign of a:

  • If a > 0 (opens up): the minimum is k, and f(x) ≥ k for all x. Range: [k, +∞)
  • If a < 0 (opens down): the maximum is k, and f(x) ≤ k for all x. Range: (−∞, k]

Example: f(x) = 2(x − 3)² − 11. Vertex is (3, −11), a = 2 > 0. Range: [−11, +∞).

What is the difference between zeros, roots, and x-intercepts?+

These three terms all refer to the same values:

  • Zero of f: a value x = a where f(a) = 0 (function perspective)
  • Root of the equation: a solution to ax² + bx + c = 0 (equation perspective)
  • x-intercept: the point (a, 0) on the graph (graph perspective)

All three phrases mean the same thing. A quadratic can have 0, 1, or 2 real zeros depending on the discriminant.

How do I write a quadratic equation given its vertex and one other point?+

Use vertex form and solve for a. Given vertex (h, k) and another point (x₀, y₀):

  1. Start with f(x) = a(x − h)² + k
  2. Substitute x = x₀ and f(x) = y₀
  3. Solve for a
  4. Write the final equation

Example: vertex (2, −3), passes through (4, 5). f(x) = a(x − 2)² − 3. Substitute (4, 5): 5 = a(4 − 2)² − 3 = 4a − 3. So 4a = 8 and a = 2.

f(x) = 2(x − 2)² − 3

How does a parabola change when you change the value of a?+

The coefficient a affects direction and width but never moves the vertex (in vertex form, h and k control position):

  • a > 0: opens upward (smile shape)
  • a < 0: opens downward (frown shape)
  • |a| > 1 (e.g., a = 3): narrows the parabola (steeper sides)
  • 0 < |a| < 1 (e.g., a = 0.5): widens the parabola (flatter sides)
  • |a| = 1: the standard width of y = x²

Compare f(x) = 5x² (very narrow) with g(x) = 0.1x² (very wide). Both open upward and have vertex at the origin, but their widths are dramatically different.

How do I find the vertex form of y = x² − 6x + 5?+

Complete the square. Here a = 1 so no initial factoring is needed:

  1. y = x² − 6x + 5
  2. Half of −6 is −3, squared is 9. Add and subtract 9:
  3. y = x² − 6x + 9 − 9 + 5
  4. y = (x − 3)² − 4

Vertex form: y = (x − 3)² − 4. Vertex: (3, −4). Opens up. Minimum is −4.

Verify: x-intercepts from factoring: (x − 5)(x − 1) = 0 → x = 1 and x = 5. The midpoint 1 and 5 is (1 + 5) / 2 = 3. This confirms the axis of symmetry at x = 3. ✓

How do I use the quadratic formula and when should I use it?+

The quadratic formula solves ax² + bx + c = 0 for any values of a, b, c:

x = (−b ± √(b² − 4ac)) / (2a)

Use the quadratic formula when:

  • The quadratic does not factor neatly over the integers
  • You need exact decimal answers
  • The discriminant is not a perfect square

Try factoring first when:

  • The leading coefficient is 1 and the constant term has small factors
  • You can spot the factors quickly (e.g., x² − 9 = (x − 3)(x + 3))

Example: solve 3x² − 7x + 2 = 0. a = 3, b = −7, c = 2. D = 49 − 24 = 25. x = (7 ± 5) / 6. So x = 2 or x = 1/3.

How do I solve a quadratic inequality like x² − 5x + 6 < 0?+

Use the graph or sign chart method:

  1. Find zeros by solving the related equation x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3
  2. The zeros divide the number line into three intervals: x < 2, 2 < x < 3, and x > 3.
  3. Test a point in each interval. Try x = 0: 0 − 0 + 6 = 6 > 0 (positive). Try x = 2.5: 6.25 − 12.5 + 6 = −0.25 < 0 (negative). Try x = 4: 16 − 20 + 6 = 2 > 0 (positive).
  4. We want < 0 (negative), which is the middle interval.

Solution: 2 < x < 3 (or in interval notation, (2, 3)).

Graphically: the parabola opens up (a = 1 > 0) and is below the x-axis between the two zeros.

How is completing the square used to derive the quadratic formula?+

The quadratic formula IS completing the square, done once in full generality. Starting from ax² + bx + c = 0:

  1. Subtract c: ax² + bx = −c
  2. Divide by a: x² + (b/a)x = −c/a
  3. Add (b/(2a))² to both sides: x² + (b/a)x + (b/(2a))² = −c/a + b²/(4a²)
  4. Left side is a perfect square: (x + b/(2a))² = (b² − 4ac)/(4a²)
  5. Take square roots: x + b/(2a) = ± √(b² − 4ac) / (2a)
  6. Solve for x: x = (−b ± √(b² − 4ac)) / (2a)

This is why completing the square is so fundamental: it proves the formula that solves every quadratic equation.

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Quadratic functions are part of a broader study of polynomial functions in precalculus. Here are the most closely related topics.

Precalculus Study Guide

Full chapter-by-chapter guide following Stewart Precalculus. Start here for context.

Polynomial Functions

Extend beyond degree 2: degree 3, 4, and higher. Zeros, multiplicity, end behavior, graphing.

Completing the Square

Deep dive into the completing-the-square technique with additional practice problems.

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