How to Find Range of Quadratic Function

Finding the range of a quadratic function is an important concept in algebra that helps us understand the set of possible output values (y-values) that the equation can produce. A quadratic equation typically has the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0.

To determine the range of a quadratic function, we need to analyze its graph, which is a parabola. The direction in which the parabola opens depends on the sign of the coefficient aaa. If aaa is positive, the parabola opens upward, and if aaa is negative, the parabola opens downward.

What is a Quadratic Function?

A quadratic equation is a second-degree polynomial equation in a single variable x with the standard form:

f(x) = ax² + bx + c

Where,

• a, b and c are constants,
• a ≠ 0 (if a = 0, the equation is linear, not quadratic).

General Solution of Quadratic Equation

The general solution to a quadratic equation can be found using the quadratic formula:

x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}

This formula provides two solutions, corresponding to the plus and minus signs before the square root, which are the roots of the quadratic equation. These roots can be real or complex numbers, depending on the discriminant b² − 4ac:

• If b² − 4ac > 0, there are two distinct real roots.
• If b² − 4ac = 0, there is one real root (a repeated root).
• If b² − 4ac < 0, there are two complex roots.

Definition of Range

Range of a function refers to the set of all possible output values (y-values) that the function can produce. In other words, it represents all the values that the dependent variable (typically y) can take as the independent variable (typically x) varies over its domain.

For a given function f: X → Y, where X is the domain of f and Y is the codomain, the range of f is the set of all y ∈ Y for which there exists at least one x ∈ X such that y = f(x). Mathematically, it is expressed as:

\text{Range}(f) = \{ y \in Y \mid \exists x \in X, \, y = f(x) \}

Methods to Find the Range for Quadratic Function

There are various methods to find range of any quadratic equation, some of these methods are:

• Using the Vertex Form
• Using Completing the Square Method
• Using Calculus (Derivatives)

Let's discuss these methods in detail as follows:

Using the Vertex Form

The vertex form of a quadratic equation is given by: y = a(x − h)² + k where (h, k) is the vertex of the parabola and vertex is either smallest or greatest value as the output of any parabolic graph.

We can use this concepts as following steps to find the range of quadratic function

Step 1: Find the Vertex of parabola i.e., (h, k) using vertex form of quadratic equation.

Step 2: Determine the Direction of the Parabola.

• If a > 0, the parabola opens upwards, and k is the minimum value.
• If a < 0, the parabola opens downwards, and k is the maximum value.

Step 3: State the range using following results.

• If the parabola opens upwards, the range is [k, ∞).
• If the parabola opens downwards, the range is (−∞, k].

Completing the Square Method

Completing the square is a method used to convert a quadratic equation into a form that makes it easier to determine key features such as the vertex and the range. Here’s a step-by-step guide on how to complete the square for a quadratic equation and use it to find the range.

Consider the quadratic equation in the standard form: y = ax² + bx + c

Isolate the quadratic and linear terms: y = ax² + bx + c

Factor out the coefficient of x² from the quadratic and linear terms: y = a(x^2 + \frac{b}{a}x) + cy

Complete the square method:

• Take the coefficient of x, divide by 2, and square it:
  • \left(\frac{\frac{b}{a}}{2}\right)^2 = \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}
• Add and subtract this square inside the parentheses:
  • y = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c
• Simplify inside the parentheses:
  • y = a\left(\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right) + c
• Distribute a:
  • y = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c

Simplify the equation:

Combine the constants: y = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)

Let h = -\frac{b}{2a} and k = c - \frac{b^2}{4a}: y = a(x - h)^2 + k

Using Calculus (Derivatives)

For a more advanced approach, calculus can be used to find the range by identifying critical points and the behavior of the function.

• Find the First Derivative: Compute the derivative of the quadratic function, f′(x) = 2ax+b
• Set the Derivative to Zero: Solve 2ax + b = 0 to find the critical point, which corresponds to the x-coordinate of the vertex.
• Second Derivative Test: Compute the second derivative, f′′(x)=2a
  • If f′′(x) > 0, the critical point is a minimum.
  • If f′′(x) < 0, the critical point is a maximum.
• Determine the Range: Use the critical point and the direction of the parabola to state the range.

Read More,

• Quadratic Equation
• Quadratic Formula
• Nature of Roots
• Derivatives
• Derivative Test
• Second Derivative Test
• Domain and Range of Function
• Domain and Range of Trigonometric Functions

Examples of Finding the Range of a Quadratic Equation

Example 1: Find range for quadratic equation: y = 2x² + 8x + 6

Solution:

y = 2(x² + 4x) + 6

⇒ y = 2(x² + 4x + 4 - 4) + 6

⇒ y = 2((x + 2)² - 4) + 6

⇒ y = 2(x + 2)² - 8 + 6

⇒ y = 2((x + 2)² - 2

The vertex form is y = 2((x + 2)² - 2

The vertex is (2, -2).

Since a = 2 > 0, the parabola opens upwards.

The range is [-2, ∞).

Example 2: Find the range of Quadratic Equation: y = -3x² + 6x + 1

Solution:

The x-coordinate of the vertex: x = -\frac{b}{2a} = -\frac{6}{2(-3)} = 1.

Substitute x = 1 into the equation to find the y-coordinate:

y = -3(1)² + 6(1) 1

⇒ y = -3 + 6 1

⇒ y = 2

The vertex is (1, 2).

Since a = -3 < 0, the parabola opens downwards.

The range is (-∞, 2].

Example 3: Find range of Quadratic Equation: y = x² + 4x + 5.

Solution:

y = (x² + 4x + 4 - 4) + 5

⇒ y = ((x + 2)² - 4) + 5

⇒ y = (x + 2)² + 1

The vertex form is y = (x + 2)² + 1.

The vertex is (-2, 1).

Since a = 1 > 0, the parabola opens upwards.

The range is [1, ∞).

Example 4: Find range of Quadratic Equation: y = x² - 4x + 3

Solution:

Differentiate the function with respect to x:

\frac{dy}{dx} = \frac{d}{dx} (x^2 - 4x + 3) = 2x - 4

To find the critical points, set the first derivative equal to zero:

2x - 4 = 0

⇒ 2x = 4

⇒ x = 2

To determine whether this critical point is a minimum or maximum, examine the second derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx} (2x - 4) = 2

Since the second derivative \frac{d^2y}{dx^2} = 2 is positive, the function has a minimum at x = 2.

Substitute x = 2 back into the original function to find the corresponding y-value:

y = (2)² - 4(2) + 3

⇒ y = 4 - 8 + 3

⇒ y = -1

Since the parabola opens upwards (as indicated by the positive second derivative), the minimum value of y is -1 and the function increases to infinity as x moves away from the vertex.

Therefore, the range of the quadratic function y = x² - 4x + 3 is: y ≥ -1

The range of the quadratic equation y = x² - 4x + 3 is: [-1, ∞)

Practice Problems on Range of a Quadratic Equation

Problem 1: Given the quadratic equation y = 3x² − 6x + 2, find the range by completing the square.

Problem 2: Given the quadratic equation y = −2x² + 4x + 5, determine the range using the vertex form.

Problem 3: Given the quadratic equation y = x² + 2x − 3, use the standard form to find the range.

Problem 4: Given the quadratic equation y = −x² + 8x − 10, find the range by identifying the vertex and the direction of the parabola.

Problem 5: Given the quadratic equation y = 2x² + 4x + 1, determine the range using the graphical method.