Table of Contents

Last modified on October 9th, 2024

A parabola has a distinct ‘U’ shaped curve that represents the graph of a quadratic function y = ax^{2} + bx + c. Here, a, b, and c are constants, and x is a variable. The shape of the parabolic curve depends on the value of the coefficient a.

- If a > 0, the parabola opens upward
- If a < 0, the parabola opens downward

The domain of a function is the set of all possible input values (generally represented by x) for which the function is defined. Since a parabola is a polynomial function, it is defined for all real numbers. This means there are no restrictions on the values that x can take. Therefore, the domain of a parabola is written in interval notation as (-∞, ∞). This implies that when we substitute any real number for x in the equation y = ax^{2} + bx + c, it will give the value of y.

The range of a function is the set of all possible output values (generally represented by y) that the function can produce. Unlike the domain, the range of a parabola depends on its orientation and the position of its vertex.

To determine the range, we need to know the **vertex** of the parabola and whether it opens upward or downward.

**For Parbola that Opens Upwards (a > 0)**

- The vertex is the minimum point on the curve
- The range starts at the y-coordinate of the vertex and extends to positive infinity.

If the vertex is at (h, k), then the range is in interval notation as [k, ∞), which means y can take any value greater than or equal to k.

**For Parbola that Opens Downwards (a < 0)**

- The vertex is the maximum point on the curve
- The range starts from negative infinity and extends to the y-coordinate of the vertex.

If the vertex is at (h, k), then the range is in interval notation as (-∞, k). This indicates that y can take any value less than or equal to k.

Let us find the domain and range of the Parabola y = x^{2} – 4x + 3

Given, y = x^{2} – 4x + 3 …..(i)

Since the domain of the quadratic equation is all real values. Thus, the domain is x ∈ (-∞, ∞)

Since the coefficient of x^{2} is 1 (positive), the parabola opens upwards.

To find the range, we first complete the square to express the quadratic in the vertex form.

From the equation (i),

y = x^{2} – 4x + 3

⇒ y = (x^{2} – 4x + 4) + 3 – 4

⇒ y = (x – 2)^{2} – 1 …..(ii)

As we know, the equation of the parabola in the vertex form is y = a(x – h)^{2} + k …..(iii)

Now, comparing the equations (ii) and (iii), we have

h = 2 and k = -1

Thus, the vertex is at (2, -1)

Thus, the range of the parabola is y ∈ (-1, ∞)

**Determine the domain and range of the parabola given by the equation:****y = -2x ^{2} + 8x – 3**

Solution:

Given, y = -2x^{2} + 8x – 3 …..(i)

Since the domain of the quadratic equation is all real values. Thus, the domain is x ∈ (-∞, ∞)

Since the coefficient of x^{2} is -2 (negative), the parabola opens downwards.

To find the range, we first complete the square to express the quadratic in the vertex form.

From the equation (i),

y = -2x^{2} + 8x – 3

⇒ y = -2(x^{2} – 4x + 4) – 3 + 8

⇒ y = -2(x – 2)^{2} + 5 …..(ii)

As we know, the equation of the parabola in the vertex form is y = a(x – h)^{2} + k …..(iii)

Now, comparing the equations (ii) and (iii), we have

h = 2 and k = 5

Thus, the range of the parabola is y ∈ (-∞, 5)

**Calculate the domain and range of y = x ^{2} + 6x + 8**

Solution:

Given, y = x^{2} + 6x + 8 …..(i)

Since the domain of the quadratic equation is all real values. Thus, the domain is x ∈ (-∞, ∞)

Since the coefficient of x^{2} is 1 (positive), the parabola opens upwards.

To find the range, we first complete the square to express the quadratic in the vertex form.

From the equation (i),

y = x^{2} + 6x + 8

⇒ y = (x^{2} + 6x + 9) + 8 – 9

⇒ y = (x + 3)^{2} – 1 …..(ii)

As we know, the equation of the parabola in the vertex form is y = a(x – h)^{2} + k …..(iii)

Now, comparing the equations (ii) and (iii), we have

h = -3 and k = -1

Thus, the range of the parabola is y ∈ (-1, ∞)

**Find the domain and range of the parabola y = -x ^{2} – 4x – 1**

Solution:

Given, y = -x^{2} – 4x – 1 …..(i)

Since the domain of the quadratic equation is all real values. Thus, the domain is x ∈ (-∞, ∞)

Since the coefficient of x^{2} is -1 (negative), the parabola opens downwards.

To find the range, we first complete the square to express the quadratic in the vertex form.

From the equation (i),

y = -x^{2} – 4x – 1

⇒ y = -(x^{2} + 4x + 4) – 1 + 4

⇒ y = -(x + 2)^{2} + 3 …..(ii)

As we know, the equation of the parabola in the vertex form is y = a(x – h)^{2} + k …..(iii)

Now, comparing the equations (ii) and (iii), we have

h = -2 and k = 3

Thus, the range of the parabola is y ∈ (-∞, 3)

**What is the domain and range of the function y = (x – 2) ^{2} + 3?**

Solution:

Given, y = (x – 2)^{2} + 3 …..(i)

The domain is x ∈ (-∞, ∞)

As we know, the equation of the parabola in the vertex form is y = a(x – h)^{2} + k …..(ii)

Comparing the equations (i) and (ii), we have

a = 1 (> 0), indicating the parabola opens upwards. Also, h = 2 and k = 3

Thus, the range of the parabola is y ∈ (3, ∞)

**Problem – **Finding the domain and range from the **GRAPH OF A PARABOLA**

**Find the domain and range of the parabola graphed below.**

Solution:

Here, the parabola opens upwards in the graph, and its vertex is at (-2, -3)

Thus, the domain is x ∈ (-∞, ∞) and the range is y ∈ (-3, ∞)

**Find the domain and range of the parabola graphed below.**

Solution:

Here, the parabola opens downwards in the graph, and its vertex is at (1, 0)

Thus, the domain is x ∈ (-∞, ∞) and the range is y ∈ (-∞, 0)

Last modified on October 9th, 2024