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Algebraic functions

Quadratic Function: Graph, Vertex, Derivative and Integral

The quadratic function f(x) = ax² + bx + c: parabola graph, vertex formula, roots, derivative 2ax + b, integral and worked calculus examples.

Algebraic functions Updated 3 min read

At a glance

f(x) = ax^2 + bx + c, \quad a \ne 0
Domain
(-\infty, \infty)
Range
\left[f\left(-\frac{b}{2a}\right), \infty\right) if a > 0
Derivative
2ax + b
Integral
\frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + C
Vertex
x = -\frac{b}{2a}
Roots
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
On this page
  1. Formula and forms
  2. Graph of a quadratic function
  3. Derivative of a quadratic
  4. Integral of a quadratic
  5. Worked examples
  6. Common mistakes
  7. How it connects

A quadratic function is a polynomial of degree two. Its graph is a parabola, a symmetric U-shaped curve with a single turning point called the vertex. Quadratics model projectile height, area and profit, and they are the first curves where derivatives reveal something you cannot see from a line.

Formula and forms

f(x) = ax^2 + bx + c, \qquad a \ne 0
Form Equation Shows
Standard ax^2 + bx + c y-intercept c
Vertex a(x - h)^2 + k vertex (h, k)
Factored a(x - r_1)(x - r_2) roots r_1, r_2

Graph of a quadratic function

Graph of the parabola y = x² - 2x - 3 and its derivative y = 2x - 2
The vertex is at (1, -4), exactly where the derivative crosses zero. The roots are x = -1 and x = 3.

Key features:

  • Direction. Opens up if a > 0 (vertex is a minimum), down if a < 0 (vertex is a maximum).
  • Axis of symmetry. The vertical line x = -\frac{b}{2a}.
  • Roots. From the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. The discriminant b^2 - 4ac tells you whether there are two, one or no real roots.
  • Range. \left[k, \infty\right) if a > 0 and \left(-\infty, k\right] if a < 0, where k is the vertex height.

Derivative of a quadratic

By the power rule and the sum rule:

f'(x) = 2ax + b, \qquad f''(x) = 2a

The derivative of a quadratic is a linear function. It is zero at exactly one point, x = -\frac{b}{2a}, which is the vertex. The second derivative is the constant 2a, so the parabola is concave up everywhere when a > 0 and concave down everywhere when a < 0. That is why the vertex is always the absolute extremum.

Integral of a quadratic

\int (ax^2 + bx + c)\,dx = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + C

The lowercase c is the quadratic's constant term; the uppercase C is the constant of integration.

Worked examples

Example 1: vertex with calculus. For f(x) = x^2 - 2x - 3, f'(x) = 2x - 2 = 0 gives x = 1, and f(1) = -4. The vertex is (1, -4), a minimum because a = 1 > 0.

Example 2: maximum height. A ball's height is h(t) = -5t^2 + 20t + 1 meters. Then h'(t) = -10t + 20 = 0 at t = 2, and the maximum height is h(2) = -20 + 40 + 1 = 21 meters.

Example 3: area under a parabola. Evaluate \int_0^3 (x^2 + 1)\,dx:

\left[\tfrac{x^3}{3} + x\right]_0^3 = (9 + 3) - 0 = 12

Example 4: tangent line. For f(x) = x^2 - 2x - 3 at x = 3: f(3) = 0 and f'(3) = 4, so the tangent line is y = 4(x - 3) = 4x - 12.

Common mistakes

  1. Forgetting the minus sign in -\frac{b}{2a}. For x^2 - 2x - 3, b = -2, so the vertex is at x = -\frac{-2}{2} = 1.
  2. Differentiating ax^2 as ax. The exponent comes down: \frac{d}{dx}3x^2 = 6x.
  3. Integrating x^2 as \frac{x^3}{2}. Divide by the new exponent, 3.
  4. Expanding (x - 2)^2 as x^2 + 4. The correct expansion is x^2 - 4x + 4.

How it connects

Quadratics are the simplest polynomial functions with curvature, so they are the standard first examples for optimization problems, curve sketching and area between curves. The second-degree Taylor polynomial of any smooth function is a quadratic, which is why parabolas approximate so many curves near a point.

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The power rule

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Frequently asked questions

What is the derivative of a quadratic function?

The derivative of ax^2 + bx + c is 2ax + b, found with the power rule term by term. For example, the derivative of 3x^2 - 5x + 2 is 6x - 5.

How do you find the vertex of a parabola with calculus?

Set the derivative to zero: 2ax + b = 0 gives x = -\frac{b}{2a}. Substitute that x into f to get the y-coordinate of the vertex.

How do you integrate a quadratic?

Use the reverse power rule on each term: \int (ax^2 + bx + c)\,dx = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + C.

How do you tell if a parabola opens up or down?

Look at the sign of a, the coefficient of x^2. If a > 0 it opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.

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