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Worked Examples · Example 13
Q.

Define the function f:R→Rf : \mathbb{R} \to \mathbb{R} by y=f(x)=x2y = f(x) = x^2, x∈Rx \in \mathbb{R}. Complete the table given below by using this definition. What is the domain and range of this function? Draw the graph of ff.

xx−4-4−3-3−2-2−1-10011223344
y=f(x)=x2y = f(x) = x^2
Punjab PsebTextbookSubjective· 5mImportance★★★★★est
37% · 37/100 Questions
✓ Free question

This problem asks you to evaluate the squaring function f(x)=x2f(x)=x^2 at integer inputs, then state its domain and range, and sketch its graph. The key idea: squaring always gives a non‑negative output, so the range is [0,∞)[0,\infty). The completed table shows the symmetric pattern of squares, and the graph is the familiar upward‑opening parabola.

Why this approach works

Function evaluation is the simplest operation in algebra: you replace the variable xx with the given number and compute. For f(x)=x2f(x)=x^2, that means multiplying the input by itself. The table is just a list of these evaluations. Once you see the outputs, the domain (all real numbers) and range (all non‑negative reals) become obvious. The graph is a parabola — symmetric about the yy-axis because (−x)2=x2(-x)^2 = x^2 — and its shape is determined by the points you’ve just computed.


1. Fill the table

For each xx, compute y=x2y = x^2:

  • x=−4x = -4: y=(−4)2=16y = (-4)^2 = 16
  • x=−3x = -3: y=(−3)2=9y = (-3)^2 = 9
  • x=−2x = -2: y=(−2)2=4y = (-2)^2 = 4
  • x=−1x = -1: y=(−1)2=1y = (-1)^2 = 1
  • x=0x = 0: y=02=0y = 0^2 = 0
  • x=1x = 1: y=12=1y = 1^2 = 1
  • x=2x = 2: y=22=4y = 2^2 = 4
  • x=3x = 3: y=32=9y = 3^2 = 9
  • x=4x = 4: y=42=16y = 4^2 = 16

The completed table:

xx−4-4−3-3−2-2−1-10011223344
y=f(x)=x2y = f(x) = x^21616994411001144991616
Tip

Notice the symmetry: f(−a)=f(a)f(-a) = f(a). This happens because (−a)2=a2(-a)^2 = a^2. So you only need to compute for non‑negative xx and mirror the results.


2. Domain

The domain is the set of all inputs for which the function is defined. Since f(x)=x2f(x)=x^2 works for every real number — you can square any real — the domain is all of R\mathbb{R}.

Domain=R=(−∞,∞)\text{Domain} = \mathbb{R} = (-\infty, \infty)


3. Range

The range is the set of all possible outputs. A square is never negative: x2≥0x^2 \ge 0 for every real xx. Also, every non‑negative number is the square of some real number (its square root). So the range is all real numbers from 00 upward.

Range=[0,∞)\text{Range} = [0, \infty)

Watch out

A common mistake is to think the range is R\mathbb{R} because the domain is R\mathbb{R}. But squaring “folds” the number line: both 22 and −2-2 give 44, and nothing gives a negative output. Always check the sign behaviour.


4. Graph of f(x)=x2f(x) = x^2

Plot the points from the table: (−4,16)(-4,16), (−3,9)(-3,9), (−2,4)(-2,4), (−1,1)(-1,1), (0,0)(0,0), (1,1)(1,1), (2,4)(2,4), (3,9)(3,9), (4,16)(4,16).

These points lie on a smooth curve called a parabola. The graph:

  • Opens upward (since the coefficient of x2x^2 is positive).
  • Has its vertex (lowest point) at (0,0)(0,0).
  • Is symmetric about the yy-axis.
  • Gets steeper as ∣x∣|x| increases.

A rough sketch:

y
|
16 -        *                 *
|
9  -    *                         *
|
4  - *                               *
|
1  -   *                           *
|
0  - *---*---*---*---*---*---*---*---*---> x
   -4  -3  -2  -1   0   1   2   3   4

(Imagine a smooth U‑shaped curve passing through all these points.)

Important

The graph of y=x2y = x^2 is the prototype parabola. Every quadratic y=ax2+bx+cy = ax^2 + bx + c is a transformation of this basic shape.


✓Final answer

The completed table shows y=16,9,4,1,0,1,4,9,16y = 16, 9, 4, 1, 0, 1, 4, 9, 16 for x=−4x = -4 to 44; the domain is R\mathbb{R}; the range is [0,∞)[0, \infty); and the graph is an upward‑opening parabola with vertex at the origin.

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